#### Answer

$$x= 0,\ x= 4$$

#### Work Step by Step

Given $$ f(x)=\frac{x^{2}}{x^{2}-4 x+8}$$
Since
\begin{align*}
f'(x)&= \frac{\frac{d}{dx}\left(x^2\right)\left(x^2-4x+8\right)-\frac{d}{dx}\left(x^2-4x+8\right)x^2}{\left(x^2-4x+8\right)^2}\\
&=\frac{2x\left(x^2-4x+8\right)-\left(2x-4\right)x^2}{\left(x^2-4x+8\right)^2}\\
&=\frac{-4x^2+16x}{\left(x^2-4x+8\right)^2}
\end{align*}
Then to find the critical points
\begin{align*}
f'(x)&=0\\
\frac{-4x^2+16x}{\left(x^2-4x+8\right)^2}&=0\\
-4x^2+16x&=0\\
4x(-x+4)&=0
\end{align*}
Hence $f(x)$ has critical points at $x= 0,\ x= 4$