Evaluating (∂U/∂T)P - (∂U/∂T)V for an ideal and van der Waals gas
Here I am going to use $U$ to refer to the total internal energy. If you want the molar internal energy ($U_\mathrm m \equiv U/n$) just divide your final result by $n$.

In another of my answers I have proven (equation $(14)$) that

$$\left(\frac{\partial U}{\partial T}\right)_{\!p} = \left(\frac{\partial U}{\partial V}\right)_{\!T}\left(\frac{\partial V}{\partial T}\right)_{\!p} + \left(\frac{\partial U}{\partial T}\right)_{\!V}$$

The proof up until there is completely general, so this applies for both an ideal and a van der Waals gas.

Furthermore here I have shown that (combining equations $(6)$ and $(8)$)

$$\left(\frac{\partial U}{\partial V}\right)_{\!T} = T\left(\frac{\partial p}{\partial T}\right)_{\!V} - p$$

Putting these together gives you

$$\begin{align}
\left(\frac{\partial U}{\partial T}\right)_{\!p} - \left(\frac{\partial U}{\partial T}\right)_{\!V} &amp;= \left(\frac{\partial V}{\partial T}\right)_{\!p}\left[ T\left(\frac{\partial p}{\partial T}\right)_{\!V} - p\right]
\end{align}$$

All that's left is to evaluate the partial derivatives. The ideal gas case is very easy; since $p = nRT/V$, $(\partial p/\partial T)_V = nR/V$ and the term in square brackets goes to zero.

For the van der Waals gas you need to work at those partial derivatives which are a bit tedious, but should be doable, especially if you employ implicit differentiation. For example, if you take the van der Waals equation of state

$$\left(p + \frac{an^2}{V^2}\right)(V - nb) = nRT$$

and differentiate with respect to $T$ throughout, keeping $V$ constant, one gets

$$\begin{align}
\left[\left(\frac{\partial p}{\partial T}\right)_{\!V} + 0\right](V - nb) + \left(p + \frac{an^2}{V^2}\right)(0) &amp;= nR \\
\left(\frac{\partial p}{\partial T}\right)_{\!V} &amp;= \frac{nR}{V - nb}
\end{align}$$

and likewise, barring any careless mistakes, I obtained

$$\left(\frac{\partial V}{\partial T}\right)_{\!p} = nR\left(p - \frac{an^2}{V^2} + \frac{2abn^3}{V^3}\right)^{-1}$$

Maybe somebody has a less tedious way (and I hope so), but this is all I can offer you myself.