The method $1$ that you suggest is a very specific case that can only be used for a general redox reaction. By this, I mean a reaction as follows for a cell having it cell notation denoted as $[\ce{N(s)|N^{y+}(aq)|| M^{x+}(aq)|M(s)}]$

\begin{array} {rlllc} \require{cancel} \ce{N^y+ + ye- &-> N } &E^\circ = E_n^\circ &\quad|\times(-x)\\ \ce{ M^x+ +xe- &-> M} &E^\circ = E_m^\circ &\quad|\times(y) \\ \hline \ce{yM^x+ + xN +\cancel{(xye-)} &-> yM + xN^y+ + \cancel{(xye-)}} &E^\circ = E_{\text{cell}}^\circ \\ \end{array}

So here, if you apply method $1$, you get $E^\circ_\text{cell} = E^\circ_m -E^\circ_n$

Let's see whether this works using method $2$ (the free energy varient). Now, we know that $\Delta G$ is additive, $\Delta G^\circ =nFE^\circ$ and that when we multiply the total reaction by a value, we also multiply the $\Delta G$ by the same number. Using these properties we find the $\Delta G$ values for the two reactions.

For just one mole of M being reduced , we get

$$\Delta G_{\text{m}}=xFE^\circ_m \tag{1}$$

For just one mole of N, we get

$$\Delta G_{\text{n}}=yFE^\circ_n \tag{2}$$

Now, according to the above cell reaction, the $\Delta G_{\mathrm {cell}} = y(\Delta G_m) + (-x)\Delta G_n$ which would be equal to

\begin{align} \Delta G_{\text{cell}} &= (-x)yFE^\circ_n + (y)xFE^\circ_m \\ &= xyF(E^\circ_m -E^\circ_n) \tag{3} \end{align}

Now, the value of $\Delta G_{\text{cell}}$ in terms of $E^\circ_\text{cell}$, we get:

$$\Delta G_{\text{cell}}=xyFE^\circ_\text{cell} \tag{4}$$

Now substituting this value of $\Delta G_\text{cell}$ into equation (3), we get

\begin{align} \cancel{(xyF)}E^\circ_\text{cell} &= \cancel{(xyF)}(E^\circ_m -E^\circ_n) \\ \implies E^\circ_\text{cell} &= E^\circ_m -E^\circ_n \end{align}

For a case where you have the same compound disproportionate, this question and its answers should be enough.

If I've understood your question properly, you are asking - in the condition $E^\circ_{\ce{M^3+/M^2+}} =\pu{4 V}$, $E^\circ_{\ce{M^3+/M}} =\pu{1 V}$, $E^\circ_{\ce{Y+/Y}} =\pu{3 V}$ - whether $\ce{M^3+}$ is a stronger oxidising agent than $\ce{Y+}$ irrespective of it going to $\ce{M \text{or} M^2+}$.

This would not be true since $\ce{Y+}$ has a higher reducing potential than $\ce{M^3+/M}$ but not $\ce{M^3+/M^2+}$, and so $\ce{M^3+}$ would be a stronger oxidising agent than $\ce{Y+}$ where it converts into $\ce{M^2+}$ but not when $\ce{M^3+}$ converts into $\ce{M}$.

The method $1$ that you suggest is a very specific case that can only be used for a general redox reaction. By this, I mean a reaction as follows for a cell having it cell notation denoted as $[\ce{N(s)|N^{y+}(aq)|| M^{x+}(aq)|M(s)}]$

\begin{array} {rlllc} \require{cancel} \ce{N^y+ + ye- &-> N } &E^\circ = E_n^\circ &\quad|\times(-x)\\ \ce{ M^x+ +xe- &-> M} &E^\circ = E_m^\circ &\quad|\times(y) \\ \hline \ce{yM^x+ + xN +\cancel{(xye-)} &-> yM + xN^y+ + \cancel{(xye-)}} &E^\circ = E_{\text{cell}}^\circ \\ \end{array}

So here, if you apply method $1$, you get $E^\circ_\text{cell} = E^\circ_m -E^\circ_n$

Let's see whether this works using method $2$ (the free energy varient). Now, we know that $\Delta G$ is additive, $\Delta G^\circ =nFE^\circ$ and that when we multiply the total reaction by a value, we also multiply the $\Delta G$ by the same number. Using these properties we find the $\Delta G$ values for the two reactions.

For just one mole of M being reduced , we get

$$\Delta G_{\text{m}}=xFE^\circ_m \tag{1}$$

For just one mole of N, we get

$$\Delta G_{\text{n}}=yFE^\circ_n \tag{2}$$

Now, according to the above cell reaction, the $\Delta G_{\mathrm {cell}} = y(\Delta G_m) + (-x)\Delta G_n$ which would be equal to

\begin{align} \Delta G_{\text{cell}} &= (-x)yFE^\circ_n + (y)xFE^\circ_m \\ &= xyF(E^\circ_m -E^\circ_n) \tag{3} \end{align}

Now, the value of $\Delta G_{\text{cell}}$ in terms of $E^\circ_\text{cell}$, we get:

$$\Delta G_{\text{cell}}=xyFE^\circ_\text{cell} \tag{4}$$

Now substituting this value of $\Delta G_\text{cell}$ into equation (3), we get

\begin{align} \cancel{(xyF)}E^\circ_\text{cell} &= \cancel{(xyF)}(E^\circ_m -E^\circ_n) \\ \implies E^\circ_\text{cell} &= E^\circ_m -E^\circ_n \end{align}

For a case where you have the same compound disproportionate, this question and its answers should be enough.

The method $1$ that you suggest is a very specific case that can only be used for a general redox reaction. By this, I mean a reaction as follows for a cell having it cell notation denoted as $[\ce{N(s)|N^{y+}(aq)|| M^{x+}(aq)|M(s)}]$

\begin{array} {rlllc} \require{cancel} \ce{N^y+ + ye- &-> N } &E^\circ = E_n^\circ &\quad|\times(-x) &(\text{oxidation})\\ \ce{ M^x+ +xe- &-> M} &E^\circ = E_m^\circ &\quad|\times(y) &(\text{reduction)} \\ \hline \ce{yM^x+ + xN +\cancel{(xye-)} &-> yM + xN^y+ + \cancel{(xye-)}} &E^\circ = E_{\text{cell}}^\circ &&\quad\text{(Cell reaction)} \\ \end{array}

So here, if you apply method $1$, you get $E^\circ_\text{cell} = E^\circ_m -E^\circ_n$

Let's see whether this works using method $2$ (the free energy varient). Now, we know that $\Delta G$ is additive, $\Delta G^\circ =nFE^\circ$ and that when we multiply the total reaction by a value, we also multiply the $\Delta G$ by the same number. Using these properties we find the $\Delta G$ values for the two reactions.

For just one mole of M being reduced , we get

$$\Delta G_{\text{m}}=xFE^\circ_m \tag{1}$$

For just one mole of N, we get

$$\Delta G_{\text{n}}=yFE^\circ_n \tag{2}$$

Now, according to the above cell reaction, the $\Delta G_{\mathrm {cell}} = y(\Delta G_m) + (-x)\Delta G_n$ which would be equal to

\begin{align} \Delta G_{\text{cell}} &= (-x)yFE^\circ_n + (y)xFE^\circ_m \\ &= xyF(E^\circ_m -E^\circ_n) \tag{3} \end{align}

Now, the value of $\Delta G_{\text{cell}}$ in terms of $E^\circ_\text{cell}$, we get:

$$\Delta G_{\text{cell}}=xyFE^\circ_\text{cell} \tag{4}$$

Now substituting this value of $\Delta G_\text{cell}$ into equation (3), we get

\begin{align} \cancel{(xyF)}E^\circ_\text{cell} &= \cancel{(xyF)}(E^\circ_m -E^\circ_n) \\ \implies E^\circ_\text{cell} &= E^\circ_m -E^\circ_n \end{align}

For a case where you have the same compound disproportionate, this question and its answers should be enough.

The method $1$ that you suggest is a very specific case that can only be used for a general redox reaction. By this, I mean a reaction as follows for a cell having it cell notation denoted as $[\ce{N(s)|N^{y+}(aq)|| M^{x+}(aq)|M(s)}]$

\begin{array} {rlllc} \require{cancel} \ce{N^y+ + ye- &-> N } &E^\circ = E_n^\circ &\quad|\times(-x)\\ \ce{ M^x+ +xe- &-> M} &E^\circ = E_m^\circ &\quad|\times(y) \\ \hline \ce{yM^x+ + xN +\cancel{(xye-)} &-> yM + xN^y+ + \cancel{(xye-)}} &E^\circ = E_{\text{cell}}^\circ \\ \end{array}

So here, if you apply method $1$, you get $E^\circ_\text{cell} = E^\circ_m -E^\circ_n$

Let's see whether this works using method $2$ (the free energy varient). Now, we know that $\Delta G$ is additive, $\Delta G^\circ =nFE^\circ$ and that when we multiply the total reaction by a value, we also multiply the $\Delta G$ by the same number. Using these properties we find the $\Delta G$ values for the two reactions.

For just one mole of M being reduced , we get

$$\Delta G_{\text{m}}=xFE^\circ_m \tag{1}$$

For just one mole of N, we get

$$\Delta G_{\text{n}}=yFE^\circ_n \tag{2}$$

Now, according to the above cell reaction, the $\Delta G_{\mathrm {cell}} = y(\Delta G_m) + (-x)\Delta G_n$ which would be equal to

\begin{align} \Delta G_{\text{cell}} &= (-x)yFE^\circ_n + (y)xFE^\circ_m \\ &= xyF(E^\circ_m -E^\circ_n) \tag{3} \end{align}

Now, the value of $\Delta G_{\text{cell}}$ in terms of $E^\circ_\text{cell}$, we get:

$$\Delta G_{\text{cell}}=xyFE^\circ_\text{cell} \tag{4}$$

Now substituting this value of $\Delta G_\text{cell}$ into equation (3), we get

\begin{align} \cancel{(xyF)}E^\circ_\text{cell} &= \cancel{(xyF)}(E^\circ_m -E^\circ_n) \\ \implies E^\circ_\text{cell} &= E^\circ_m -E^\circ_n \end{align}

For a case where you have the same compound disproportionate, this question and its answers should be enough.