= Solution
Start with the fundamental relation at fixed material amount, $dU=T\,dS-p\,dV$, and expand <entropy> in independent variables $T,V$:
$$
dS=\left(\frac{\partial S}{\partial T}\right)_VdT
+\left(\frac{\partial S}{\partial V}\right)_TdV
=\frac{C_V}{T}dT+\left(\frac{\partial p}{\partial T}\right)_VdV.
$$
The last equality uses the supplied <Maxwell relation>. Consequently
$$
dU=C_V\,dT+\left[T\left(\frac{\partial p}{\partial T}\right)_V-p\right]dV.
$$
For the <excluded-volume ideal gas>, $(\partial p/\partial T)_V=R/(V-b)=p/T$, so the bracket vanishes:
$$
\boxed{\left(\frac{\partial U}{\partial V}\right)_T=0,
\qquad U=U(T),\qquad dU=C_V\,dT.}
$$
Once $U$ depends only on <temperature>, $C_V=dU/dT$ cannot depend on <volume (thermodynamics)> either. Its stipulated <temperature> independence therefore makes it a constant, giving $U=C_VT+U_*$ with arbitrary <energy> reference $U_*$. <Excluded volume> changes the <pressure> without adding the volume-dependent attractive <internal energy> that a more general interacting-gas model might have.
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