Answer
At $x=e^{-1/2}$, the function has an absolute minimum of $\displaystyle \frac{-1}{2e}$
Work Step by Step
The domain is $(0,+\infty)$.
The graph crosses the x-axis at $x=1.$
When $x\rightarrow 0^{+},$ observing a few function values
$\left[\begin{array}{l}
f(0.1)\approx-0.023\\
f(0.001)\approx-6.9\times 10^{-6}
\end{array}\right],$
we note that the graph is below the x-axis, approaching the origin.
$\displaystyle \lim_{x\rightarrow\infty}f(x)=\infty,\qquad $(there are no absolute maxima)
Find critical points:
$f'(x)=[x^{2}]'\displaystyle \ln x+x^{2}\cdot[\ln x]'=2x\ln x+x^{2}\cdot\frac{1}{x}=x(2\ln x+1)$
$f'(x)=0$ when $\left[\begin{array}{l}
2\ln x+1=0\\
2\ln x=-1\\
\ln x=-\frac{1}{2}\\
x=e^{-1/2}\approx 0.60653
\end{array}\right]$
Evaluate at $x=e^{-1/2},\quad $
$f(e^{-1/2})=(e^{-1/2})^{2}\displaystyle \cdot(-\frac{1}{2})=\frac{-1}{2e}\approx-0.1839$
At $x=e^{-1/2}$, the function has an absolute minimum of $\displaystyle \frac{-1}{2e}$
