Answer
$$y'=\frac{-\pi\sin (\cos \sqrt{\sin \pi x} ) [\sin \sqrt{\sin \pi x}]\cos \pi x}{ \sqrt{\sin \pi x}}$$
Work Step by Step
Given $$ y = \sin^2 (\cos \sqrt{\sin \pi x} ) $$
Then
\begin{align*}
y'& =2\sin (\cos \sqrt{\sin \pi x} ) [-\sin \sqrt{\sin \pi x}]\frac{1}{2 \sqrt{\sin \pi x}}(\pi\cos \pi x)\\
&=\frac{-\pi\sin (\cos \sqrt{\sin \pi x} ) [\sin \sqrt{\sin \pi x}]\cos \pi x}{ \sqrt{\sin \pi x}}
\end{align*}
