We have,
$f(x)=x \log x$
$f'(x)=1+\log x$
For maxima or minima, put $f^{\prime}(x)=0$
$\therefore 1+\log x=0 $
$\Rightarrow \log x=-1 $
$\Rightarrow x=e^{-1}$
Now,
$f''(x)=\frac{1}{x}$
$\Rightarrow f''\left(e^{-1}\right)=\frac{1}{e^{-1}}=e>0$
$\therefore f(x)$ is decreasing. Hence, minimum value of
$f(x)$ at $x=e^{-1}$ is
$f\left(e^{-1}\right)=e^{-1} \log e^{-1}=\frac{1}{e}(-\log e)=-\frac{1}{e}$