Answer
$$ y'= 2x(\cos^2(x^2)-\sin^2(x^2)).$$
Work Step by Step
Recall that $(\sin x)'=\cos x$.
Recall that $(\cos x)'=-\sin x$.
Recall that $(x^n)'=nx^{n-1}$
Since $ y=\sin(x^2) \cos(x^2)$, then by the chain and product rules, the derivative $ y'$ is given by
$$ y'= 2x\cos(x^2) \cos(x^2)-2x\sin(x^2) \sin(x^2)=2x(\cos^2(x^2)-\sin^2(x^2)).$$
