Given, curve is $y=x e^{x}$
$\Rightarrow \frac{d y}{d x}=e^{x}+x e^{x}$
For maximum and minimum, put $\frac{d y}{d x}=0$
$\Rightarrow e^{x}(1+x)=0$
$\Rightarrow x=-1$
Now, $\frac{d^{2} y}{d x^{2}}=2 e^{x}+x e^{x}$
At $x=-1, \frac{d^{2} y}{d x^{2}}=e^{-1}(2-1)>0$
Hence, $x=-1$ is a point of minimum.