Answer
$x=0,\ \ x=-6,\ x=6$
Work Step by Step
Given
$$y=\frac{x}{\left(x^{2}-4\right)^{1 / 3}}$$
Since
\begin{align*}
y'&=\frac{\frac{d}{dx}\left(x\right)\left(x^2-4\right)^{\frac{1}{3}}-\frac{d}{dx}\left(\left(x^2-4\right)^{\frac{1}{3}}\right)x}{\left(\left(x^2-4\right)^{\frac{1}{3}}\right)^2}= \frac{x^2-12}{3\left(x^2-4\right)^{\frac{4}{3}}} \\
y''&=\frac{\frac{d}{dx}\left(x^2-12\right)\left(x^2-4\right)^{\frac{4}{3}}-\frac{d}{dx}\left(\left(x^2-4\right)^{\frac{4}{3}}\right)\left(x^2-12\right)}{3\left(\left(x^2-4\right)^{\frac{4}{3}}\right)^2}= \frac{2x\left(-x^2+36\right)}{9\left(x^2-4\right)^{\frac{7}{3}}}
\end{align*}
Then $f''(x)=0$ for $x=0,\ \ x=\pm 6$. We note that $y''$ changes from negative to positive around these critical points.
Hence, there exist inflection points at $x=0,\ \ x=-6,\ x=6$
