#### Answer

$x=2,\ \ x=2/3$
$ f(x)$ has a local maxima at $x= 2/3$ and has a local minima at $x= 2$

#### Work Step by Step

Given $$f(x)=x^{3}-4 x^{2}+4 x $$
Since
\begin{align*}
f'(x) &= 3x^2-8x+4\\
f''(x)&= 6x-8
\end{align*}
Since $f(x)$ has a critical point when $f'(x)= 0$
\begin{align*}
f'(x)&=0\\
3x^2-8x+4&=0\\
(3x-2)(x-2)&=
0\end{align*}
Then $f(x)$ has critical points at $x= 2,\ x=2/3$, and
\begin{align*}
f''(3/2)&= -4<0\\
f''(2)&=4>0
\end{align*}
Then $ f(x)$ has a local maxima at $x= 2/3$ and has a local minima at $x= 2$.