Answer
$$ f'(x)= \sin (2x) e^{\sin^2 x}.$$
Work Step by Step
Recall that $(e^x)'=e^x$
Recall that $(\sin x)'=\cos x$.
Recall that $2\sin x \cos x=\sin 2x$
Since $ f(x)=e^{\sin^2 x}$, then the derivative, using the chain rule, is given by
$$ f'(x)=e^{\sin^2 x} (\sin^2 x)'=(2\sin x\cos x)e^{\sin^2 x}=\sin (2x) e^{\sin^2 x}.$$
