Answer
$x=\pm 1, x=0, x=\pm\sqrt {\frac{2}{3}}$
Work Step by Step
$f'(x)=\frac{d}{dx}(x^{2})\times\sqrt {1-x^{2}}+x^{2}\times\frac{d}{dx}(\sqrt {1-x^{2}})$
$=2x\sqrt {1-x^{2}}+x^{2}\times\frac{1}{2\sqrt {1-x^{2}}}\times\frac{d}{dx}(1-x^{2})$
$=2x\sqrt {1-x^{2}}-\frac{x^{3}}{\sqrt {1-x^{2}}}$
$f'(x)$ does not exist if $\sqrt {1-x^{2}}=0$, since the denominator cannot be 0.
$\sqrt {1-x^{2}}=0\implies x= \pm 1$
$f'(x)=0\implies 2x\sqrt {1-x^{2}}=\frac{x^{3}}{\sqrt {1-x^{2}}}$
$\implies 2x(1-x^{2})=x^{3}$
$\implies x= 0$ or $x= \pm \sqrt {\frac{2}{3}}$
The critical points of $f$ are those $x$ for which $f'(x)$ does not exist or $f'(x)=0$. That is, the critical points are $x=\pm 1, x=0, x=\pm\sqrt {\frac{2}{3}}$
