Answer
$ x=1$, local minima at $ x=1$
Work Step by Step
To find the critical point, we put $ f'(x)=0$, so we have
$$ f(x)=\frac{e^{x}}{x}\Longrightarrow f'(x)=\frac{xe^x-e^x}{x^2}=0,$$
then $ xe^{x}=e^x $ and hence $ x=1$. So the critical point is $ x=1$.
Moreover, we have $$ f''(x)=\frac{x^2(xe^x)-2x(xe^x-e^x)}{x^4} =\frac{x^3-2x(x-1)}{x^4}e^x $$
then $$ f''(1)=e>0 $$
then $ f(x) $ has a local minima at $ x=1$.
