Why is $\left.\frac{\partial C_V}{\partial V}\right|_T$ different in these derivations? I want to show that $\delta q$ is not an exact differential.
Starting from $dE = \delta q - pdV$ and because $E := E(V, T)$ is a state function, which allows to express the exact differential as
$$
dE = \left.\frac{\partial E}{\partial V}\right|_T dV +\left.\frac{\partial E}{\partial T}\right|_V dT, 
$$
the two expressions can be set equal giving after rearrangement
$$
\delta q = \left[ \left.\frac{\partial E}{\partial V}\right|_T + p \right]dV + \left.\frac{\partial E}{\partial T}\right|_V dT
$$
and therefore also
$$
\delta q = \left[ \left.\frac{\partial E}{\partial V}\right|_T + p \right]dV + C_V dT.
$$
Now, by definition $\partial E/\partial T|_V = C_V$, and so
$$
\left.\frac{\partial C_V}{\partial V}\right|_T =\left[\frac{\partial}{\partial V}\left.\frac{\partial E}{\partial T}\right|_V\right]_T
$$
and because $E$ is a state function, the sequence of partial derivatives can be exchanged (according to Schwarz' theorem, while I don't understand how it works), allowing to write
$$
\left.\frac{\partial C_V}{\partial V}\right|_T =\left[\frac{\partial}{\partial T}\left.\frac{\partial E}{\partial V}\right|_T\right]_V. (*)
$$
Also, multiplying the above expression for $\delta q$ by $1/\partial V$ at constant $T$, I obtain
$$
\left.\frac{\delta q}{\partial V}\right|_T = \left[ \left.\frac{\partial E}{\partial V}\right|_T + p \right]\left.\frac{\partial V}{\partial V}\right|_T + C_V \left.\frac{\partial T}{\partial V}\right|_T
$$
where $\partial V/\partial V = 1$ and the second term on the right hand side equals $0$ because $\partial T = 0$ at constant temperature. Multiplying the remaining equation by $\partial / \partial T$ at constant $V$ gives
$$
\left[\frac{\partial}{\partial T}\left.\frac{\delta q}{\partial V}\right|_T\right]_V = \left[ \frac{\partial}{\partial T} \left(\left.\frac{\partial E}{\partial V}\right|_T + p \right)\right]_V.
$$
Now assuming $\delta q$ were exact, again the sequence of partial derivatives would not matter and I could write
$$
\left[\frac{\partial}{\partial V}\left.\frac{\delta q}{\partial T}\right|_V\right]_T = \left[ \frac{\partial}{\partial T} \left(\left.\frac{\partial E}{\partial V}\right|_T + p \right)\right]_V
$$
and by using $q=E$ since the "inner" differential on the left side is evaluated at constant volume,
$$
\left[\frac{\partial}{\partial V}\left.\frac{\partial E}{\partial T}\right|_V\right]_T = \left.\frac{\partial}{\partial V} C_V\right|_T = \left[ \frac{\partial}{\partial T} \left(\left.\frac{\partial E}{\partial V}\right|_T + p \right)\right]_V. (**)
$$
From this we find that $(*)$ and $(**)$ are different and thus the assumption must be wrong and therefore $\delta q$ is not an exact differential.
Does this make any sense?

Exact wording from book:

Starting with $dE = \delta q - pV$, show that
a) $\delta q = C_V dT + [P+(\partial E/\partial V)_T] dV$
b) $\left(\frac{\partial C_V}{\partial V}\right)_T = \left[\frac{\partial}{\partial T} \left(\frac{\partial E}{\partial V}\right)_T\right]_V$
c) $\delta q$ is not an exact differential.

For c), the book states

If $\delta q$ were an exact differential, then by solution to a), $(\partial C_V/\partial V)_T$ would have to be equal to $[\partial /\partial T(P+(\partial E/\partial V)_T)]_V$ but it is not according to solution of b), hence $\delta q$ is not exact.

 A: I think it is important to not loose yourself in calculations. The method in your book probably starts from saying that considering an object $\delta Q$ that has the following general form:
$\delta Q = C_vdT + hdV$ say,
then it is an exact differential iif
$\left(\frac{\partial C_v}{\partial V}\right)_T = \left(\frac{\partial h}{\partial T}\right)_V$
This is also equivalent to saying that $C_v$ and $h$ can be thought of as partial derivatives of the same state function with respect to $T$ and $V$ respectively.
Now, from the first law of thermodynamics, we know that 
$C_v \equiv \left( \frac{\partial U}{\partial T}\right)_V$
hence the point is then to figure out if $h$ is the partial derivative of $U$ with respect to $V$ at fixed $T$ and obviously it is not since the latter would be $h-p$ where $p$ is the thermodynamic pressure which is not a null function.
That ends the story I believe...but please tell me if you don't trust this argument, I can easily be fooled myself by circular arguments.
