Answer
$ x=0$, local minima at $ x=0$.
Work Step by Step
To find the critical point, we put $ f'(x)=0$, so we have
$$ f(x)=e^{-x}+x\Longrightarrow f'(x)=-e^{-x}+1=0,$$
then $ e^{-x}=1$ and hence $ x=0$. So the critical point is $ x=0$.
Moreover, we have $$ f''(x)=e^{-x}\Longrightarrow f''(0)=1>0$$
then $ f(x) $ has local minima at $ x=0$.
