These statements can be proved in general i.e. irrespective of whether you take your system as ideal gas or not. As you have used in the question $PV = nRT$, you inherently assuming that your system of thermodynamic interest is an ideal gas. So, let's first prove this for an ideal gas and then do the general proof.
$\textbf{Case I : Ideal Gas}$
For an ideal gas, $C_V(T)$ is independent of volume $V$, which means that $\big(\frac{\partial C_V(T)}{\partial V}\big)_T =0$. (We will prove the general relation in the next section, and this statement will be obvious from that proof. ), But on the other hand $\big(\frac{\partial (nR T/V)}{\partial T}\big)_V = \frac{nR}{V}$ by simple partial derivative. So, thus you get the proof that $\big(\frac{\partial C_V(T)}{\partial V}\big)_T \neq \big(\frac{\partial (nR T/V)}{\partial T}\big)_V$. Thus $\delta q_{rev}$ is not an exact differential.
On the other hand if you divide both sides by $T$, you still have, $\big(\frac{\partial (C_V(T)/T)}{\partial V}\big)_T = \frac{1}{T}\big(\frac{\partial C_V(T)}{\partial V}\big)_T =0$. And also on the other hand you have, $\big(\frac{\partial (nR /V)}{\partial T}\big)_V = -\frac{nR}{V^2}\big(\frac{\partial V}{\partial T}\big)_V = 0$ (obviously, as we are taking partial derivative of a constant term. Thus, by the equality of mixed partial derivatives of $\frac{\delta q_{rev}}{T}$ it becomes exact differential.
$\textbf{Case II : General System}$
For general system we have, $\delta q_{rev} = dU +PdV$. But remember we can't write, $dU = C_V(T) dT $ and $PV =nRT$. We have to write $dU = C_V(T)dT + \big(\frac{\partial U}{\partial V}\big)_T \ dV$ .
So, now we have $\delta q_{rev} = C_V(T)dT + \Big(\big(\frac{\partial U}{\partial V}\big)_T + P\Big)dV $.
Also, from the definition of Helmholtz free energy ($A = U - TS $ and $dA = -SdT -PdV$) and by using Maxwell's relation it can be shown that $\Big(\big(\frac{\partial U}{\partial V}\big)_T + P\Big) = T \big(\frac{\partial P}{\partial T}\big)_V$
Now, it can be shown properly that
$$\left(\frac{\partial C_V(T)}{\partial V}\right)_T = \frac{\partial^2 U}{\partial T \partial V} = T \left(\frac{\partial^2 P}{\partial T^2}\right)_V \qquad(= 0\text{ if } PV=nRT)\tag{i} $$
$$\left(\frac{\partial (\frac{C_V(T)}{T})}{\partial V}\right)_T = \frac{1}{T} \frac{\partial^2 U}{\partial T \partial V} = \left(\frac{\partial^2 P}{\partial T^2}\right)_V \tag{ii}$$
and for the right hand side
$$\frac{\partial}{\partial T}\Bigg(T\Big(\frac{\partial P}{\partial T}\Big)_V\Bigg)_V = T\Big(\frac{\partial^2 P}{\partial T^2}\Big)_V + \left(\frac{\partial P}{\partial T}\right)_V\tag{iii}$$
and if you divide the right side by $T$ and then check the partial derivative criterion, you will ultimately get $\Big(\frac{\partial^2 P}{\partial
T^2}\Big)_V$ on the r.h.s also (exactly same as $(ii)$ ) , But note that, $(i) \neq (iii) $. Thus $\frac{\delta q_{rev}}{T}$ will become exact differntial but $\delta q_{rev}$ will remain inexact.