Answer
Concave upward on
$$
(-\infty, -e^{-\frac{3}{2} } ), \quad (e^{-\frac{3}{2} }, \infty)
$$
concave downward on
$$
(-e^{-\frac{3}{2} } ,0) , \quad (0,e^{-\frac{3}{2} } ), $
$$
inflection point at
$$
(-e^{-\frac{3}{2} } ,-\frac{3e^{-3}}{2\ln(10)} ), \quad (e^{-\frac{3}{2} } ,-\frac{3e^{-3}}{2\ln(10)} ).
$$
Work Step by Step
$$
f(x)=x^{2} \log |x|
$$
The first derivative is
$$
\begin{aligned}
f^{\prime}(x) &=2x \log |x| +x^{2} \frac{1}{x\ln(10)}\\
&=2x \log |x| + \frac{x}{\ln(10)}
\end{aligned}
$$
and the second derivative is
$$
\begin{aligned}
f^{\prime\prime}(x) &=2\log |x| + \frac{2x}{x\ln(10)} +\frac{1}{\ln(10)}\\
&=2\log |x| + \frac{3}{\ln(10)}\\
\end{aligned}
$$
A function $f$ is concave upward on an interval if $f^{\prime\prime}(x) \gt 0$ for all $x$ in the interval, so
$$
\begin{aligned}
f^{\prime\prime}(x) &=2\log |x| + \frac{3}{\ln(10)} \gt 0 \\
\Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
2\log |x| &\gt -\frac{3}{\ln(10)} \\
\log |x| & \gt -\frac{3}{2\ln(10)} \\
\frac{\ln |x|}{\ln(10)} & \gt -\frac{3}{2\ln(10)} \\
\ln |x| & \gt -\frac{3}{2} \\
|x| & \gt e^{-\frac{3}{2} }\\
\Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
x \lt -e^{-\frac{3}{2} } \quad \text {and} & \quad x \gt e^{-\frac{3}{2} }\\
\end{aligned}
$$
so $f(x)$ is concave upward on $(-\infty, -e^{-\frac{3}{2} } ), \quad (e^{-\frac{3}{2} }, \infty)$.
Also function $f$ is concave downward on an interval if $f^{\prime\prime}(x) \lt 0$ for all $x$ in the interval, so
$$
\begin{aligned}
f^{\prime\prime}(x) &=2\log |x| + \frac{3}{\ln(10)} \lt 0 \\
\Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
2\log |x| & \lt -\frac{3}{\ln(10)} \\
\log |x| & \lt -\frac{3}{2\ln(10)} \\
\frac{\ln |x|}{\ln(10)} & \lt -\frac{3}{2\ln(10)} \\
\ln |x| & \lt -\frac{3}{2} \\
|x| & \lt e^{-\frac{3}{2} }\\
\Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
-e^{-\frac{3}{2} } & \lt x \lt e^{-\frac{3}{2} }\\
\end{aligned}
$$
But $f(x)$ is not defined at $x=0$
so $f(x)$ is concave downward on $(-e^{-\frac{3}{2} } ,0) $ and $(0,e^{-\frac{3}{2} } ) $ .
Since the sign of $f^{\prime\prime}(x)$ changes from positive to negative at $ -e^{-\frac{3}{2} } $, the graph changes from concave upward to concave downward at that point, so $(-e^{-\frac{3}{2} } , f(-e^{-\frac{3}{2} }))$ is inflection point.
Also, the sign of $f^{\prime\prime}(x)$ changes from negative to positive at $e^{-\frac{3}{2} } $, the graph changes from concave downward to concave upward at that point, so $(e^{-\frac{3}{2} } , f(e^{-\frac{3}{2} }))$ is inflection point.
Now ,we find $f( -e^{-\frac{3}{2} }), f(e^{-\frac{3}{2} })$as follows:
$$
\begin{aligned}
f(-e^{-\frac{3}{2} }) &=(-e^{-\frac{3}{2} })^{2} \log |-e^{-\frac{3}{2} }| \\
&=e^{-3} \log e^{-\frac{3}{2} }\\
&=-\frac{3e^{-3}}{2\ln(10)}
\end{aligned}
$$
and
$$
\begin{aligned}
f(e^{-\frac{3}{2} }) &=(e^{-\frac{3}{2} })^{2} \log |e^{-\frac{3}{2} }| \\
&=e^{-3} \log e^{-\frac{3}{2} }\\
&=-\frac{3e^{-3}}{2\ln(10)}
\end{aligned}
$$
Finally, we have inflection points at
$$
(-e^{-\frac{3}{2} } ,-\frac{3e^{-3}}{2\ln(10)} ), \quad (e^{-\frac{3}{2} } ,-\frac{3e^{-3}}{2\ln(10)} ).
$$
