Answer
$$\frac{\partial}{\partial p}\tan^{-1}(pq^2)=\frac{q^2}{1+p^2q^4}.$$
$$\frac{\partial}{\partial q}\tan^{-1}(pq^2)=\frac{2pq}{1+p^2q^4}.$$
Work Step by Step
Regard $q$ as constant and differentiate with respect to $p$:
$$\frac{\partial}{\partial p}\tan^{-1}(pq^2)=\frac{1}{1+(pq^2)^2}\frac{\partial}{\partial p}(pq^2)=\frac{1}{1+p^2q^4}q^2=\frac{q^2}{1+p^2q^4}.$$
Regard $p$ as constant and differentiate witth respect to $q$:
$$\frac{\partial}{\partial q}\tan^{-1}(pq^2)=\frac{1}{1+(pq^2)^2}\frac{\partial}{\partial q}(pq^2)=\frac{1}{1+p^2q^4}2pq=\frac{2pq}{1+p^2q^4}.$$
