Answer
$ x=0, -2$, $ f(x) $ has a local minimum at $ x=0$ and local maximum at $ x=-2$.
Work Step by Step
To find the critical point, we put $ f'(x)=0$, so we have
$$ f(x)=x^2 e^{x}\Longrightarrow f'(x)=2xe^{x}+x^2e^{x}=(x^2+2x)e^{x}=0,$$
then $ x(x+2)=0$ and hence $ x=0$ or $ x=-2$. So the critical points are $ x=0$ and $ x=-2$..
Moreover, we have $$ f''(x)=(2x+2 )e^{x}+(x^2+2x)e^{x}$$
then $$ f''(0)=2>0, \quad f''(-2)=-2e^{-2}<0. $$
Then $ f(x) $ has a local minimum at $ x=0$ and local maximum at $ x=-2$.
