The standard electrode potentials for the following redox couples are given as: >$E_0(\ce{Fe^{2+}/Fe}) = x~\mathrm{V}$ and $ E_0(\ce{Fe^{3+}/Fe^{2+}}) = y~\mathrm{V}$
What will be potential $E_0(\ce{Fe^{3+}/Fe})$ (in $\mathrm{V}$)?
My approach: $\ce{Fe^{3+} + e- + Fe^{2+} + 2e- -> Fe^{2+} + Fe}$
So $E_0(\ce{Fe^{3+}/Fe})=(x+y)~\mathrm{V}$
You can write down the three half reactions:
\begin{aligned} \ce{Fe^{2+} + 2e- &-> Fe} & E_1&= x\,\mathrm{V}\\ \ce{Fe^{3+} + e- &-> Fe^{2+}} & E_2&= y\,\mathrm{V}\\ \ce{Fe^{3+} + 3e- &-> Fe} & E_3&= z\,\mathrm{V} \end{aligned}
The third half-reaction is simply the sum of the first two. However, we cannot just sum $E_1 + E_2$ in order to obtain $E_3$. The problem is with the meaning of the potentials, and their relation to the stoichiometry.
An half reaction with potential $E$ in volts means that the elecrons have some energy assigned to them. The meaning of volts is joules per coulomb. The charge of one mole of electrons is 96500 C (this is Faraday's constant: 1 F = 95600C/mol).
Take a look at the first half reaction. It transfers two moles of electron per mole of reaction. This means the electrons carry
In the same fashion, the second half reaction has the electrons carrying
The third half reaction is the sum of the first two, so its electrons carry
Now you can simplify this to $2x + y = 3z$ and calculate the potential you want.
As correct as eflaschuks approach is, I found it rather confusing, because much text and few mathematics. So I am adding them here.
First of all, let us write down the Nernst Equation for an arbitrary reaction $\ce{Ox + z\cdot e- <=> Red}$: \begin{aligned} E=E^\ominus(\ce{Red/Ox}) - z^{-1}\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Red}}{\ce{Ox}} \end{aligned}
For our given system that boils down to the following reaction and their potentials: \begin{aligned} \ce{Fe^{2+} + $2\cdot$ e- &-> Fe}& E_1 &= E^\ominus(\ce{Fe/Fe^{2+}}) - \frac12\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{2+}}}& \text{(1)}\\ \ce{Fe^{3+} + $1\cdot$ e- &-> Fe^{2+}}& E_2 &= E^\ominus(\ce{Fe^{2+}/Fe^{3+}}) - \frac11\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe^{2+}}}{\ce{Fe^{3+}}}& \text{(2)}\\ \ce{Fe^{3+} + $3\cdot$ e- &-> Fe}& E_3 &= E^\ominus(\ce{Fe/Fe^{3+}}) - \frac13\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{3+}}}& \text{(3)}\\ \end{aligned}
Now you are looking for the standard potential $E^\ominus(\ce{Fe/Fe^{3+}})$, so you have to consider the potential $E_3$, which can be build from $E_1$ and $E_2$: $$E_3 = a\cdot E_1 + b\cdot E_2$$
Substituting $(1)$, $(2)$ and $(3)$ yields:
$$\begin{multline} E^\ominus(\ce{Fe/Fe^{3+}}) - \frac13\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{3+}}} = \\ a\cdot\left(E^\ominus(\ce{Fe/Fe^{2+}}) - \frac12\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{2+}}}\right) \\+ b\cdot\left(E^\ominus(\ce{Fe^{2+}/Fe^{3+}}) - \frac11\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe^{2+}}}{\ce{Fe^{3+}}}\right) \end{multline}$$
In order to obtain the standard potential the logarithmic terms have to cancel. Therfore we can split these equations: \begin{aligned} &&E^\ominus(\ce{Fe/Fe^{3+}}) &= a\cdot E^\ominus(\ce{Fe/Fe^{2+}}) + b\cdot E^\ominus(\ce{Fe^{2+}/Fe^{3+}}) \\ &&\frac13\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{3+}}} &= a\cdot\left(\frac12\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{2+}}}\right) + b\cdot\left(\frac11\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe^{2+}}}{\ce{Fe^{3+}}}\right)\\ \therefore&& \ln\frac{\ce{Fe}}{\ce{Fe^{3+}}} &= \frac{3a}{2} \ln\frac{\ce{Fe}}{\ce{Fe^{2+}}} + 3b \ln\frac{\ce{Fe^{2+}}}{\ce{Fe^{3+}}} \end{aligned}
In order for the $\ce{Fe^{2+}}$ terms to cancel it is fairly obvious, that $$a=\frac23,\\ b=\frac13,$$ therfore resulting in \begin{aligned} &&E^\ominus(\ce{Fe/Fe^{3+}}) &= \frac{2}{3}\cdot E^\ominus(\ce{Fe/Fe^{2+}}) + \frac{1}{3}\cdot E^\ominus(\ce{Fe^{2+}/Fe^{3+}}). \end{aligned}
You can also check that against experimental data (from wikipedia): \begin{array}{llrlr}\hline \text{halfreaction} &&&& \text{potential}\\\hline \ce{Fe^{2+} + $2\cdot$ e- &-> Fe} & E_1 &=& -0.44\mathrm{V} \\ \ce{Fe^{3+} + $1\cdot$ e- &-> Fe^{2+}}& E_2 &=& 0.77\mathrm{V}\\ \ce{Fe^{3+} + $3\cdot$ e- &-> Fe} & E_3 &=& -0.04\mathrm{V}\\\hline \end{array}
Your approach is a correct one.
Explanation: (I call the given two reduction half cell reactions as 1 and 2 respectively)
The required half cell reaction $\ce{Fe^{3+} -> Fe}$ could be obtained by adding reaction 1 and 2 .Thus the net $\ce{E0}$ value would be the sum of $\ce{E0}$ values of reaction 1 and 2 (as both are reduction half cell reactions.)
Hence net $\ce{E0(Fe^{3+}/Fe^{2+})} = (x + y) \, \text{V}$
