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$.
