Answer
$ x=0 $ and $ x=4$
Work Step by Step
Given $$ f(x) = \frac{x^2}{x-2}$$
Then to find the critical points
\begin{align*}
f'(x)&=0\\
\frac{(x-2 ) 2 x-x^{2}(1)}{(x-2)^{2}}&=0\\
\frac{x^{2}-4 x}{(x-2)^{2}} &=0\\
x^{2}-4 x &=0
\end{align*}
Hence the critical points are $ x=0 $ and $ x=4$
