How do I properly partially differentiate with constant $p$ in thermodynamics? I'm trying to solve the following problem:

a one component system is described by the following equations
$$U=\frac{A^2}{4}NT^2\exp \left(\frac{V^2}{N^2}\right),\qquad p=T^2f(v)$$
where $v = \frac{V}{N},\,\,A\in ℝ$.

Through clever use of symmetry of second derivatives of $S$ I got to the following equation for $p$ and $S$:
$$
p = -T^2\frac{A^2}{2}v\text{ Exp}[v^2];\qquad\qquad S =\frac{A^2}{2}NT\text{ Exp}\left[v^2\right] 
$$
I believe these are correct. Now I thought of using the definition formula
$$
C_p=T\left(\frac{\partial S}{\partial T}\right)_{p\,=\text{ const.}}
$$
and I thought to myself "hey! The functions present in $p$ are also in $S$! If I factor them out correctly, I'll get the desired dependence easily":
$$
S = -\frac{pN^2}{VT}
$$
That would be easy enough! But then I realized i could just as much "factor out" $\sqrt{p}$, kill all dependence on $T$ and arrrive at $C_p = 0$.
Why can this happen, what did I do wrong and how do I handle this correctly?
EDIT: Ok, we came to the conclusion that $V = V(T,p)$ which I didn't consider when derivating (thank you, Andrew), but now I have completely no idea how to solve this. From the equation of $p$ one can't get $V=V(T,p)$ and without it I can't get $\left(\frac{\partial V}{\partial T}\right)_p$ and solve this. How do I correctly find $C_p$?
 A: Yes we can write $S$ as a function of $V$ and $p$ but it does not follow that $(\partial S/\partial T)_p = 0$. Rather:
$$
S=S(V,p)
$$
$$
dS = \left. \frac{\partial S}{\partial p}\right|_V dp + 
 \left. \frac{\partial S}{\partial V}\right|_p dV
$$
so
$$
\left. \frac{\partial S}{\partial T}\right|_p = 
 \left. \frac{\partial S}{\partial V}\right|_p \left. \frac{\partial V}{\partial T}\right|_p
$$
The fact that we can write $S$ as a function of $p$ and $V$ does not "kill all the dependence on $T$" as you put it. The dependence on $T$ is still there because $V$ depends on $T$ (at any given pressure).
A: I have some hints for you on how to solve for Cp.
$C_p$ is defined as $$C_p=\frac{1}{N}\left(\frac{\partial H}{\partial T}\right)_P$$Enthalpy H is defined as $$H=U+PV$$You have equations for U and P, so you know H = H(T,V).  Using this, the differential of H is $$dH=\left(\frac{\partial H}{\partial T}\right)_VdT+\left(\frac{\partial H}{\partial V}\right)_TdV$$So, $$\left(\frac{\partial H}{\partial T}\right)_P=\left(\frac{\partial H}{\partial T}\right)_V+\left(\frac{\partial H}{\partial V}\right)_T\left(\frac{\partial V}{\partial T}\right)_P$$The differential of P in terms of dV and dT is:  $$dP=\left(\frac{\partial P}{\partial T}\right)_VdT+\left(\frac{\partial P}{\partial V}\right)_TdV$$Since, at constant pressure, dP=0, this allows you to express $\left(\frac{\partial V}{\partial T}\right)_P$ in terms of $\left(\frac{\partial P}{\partial T}\right)_V$ and $\left(\frac{\partial P}{\partial V}\right)_T$
