Answer
Stationary: $x=\frac{3}{4}$ and $x=0$
Critical: $x=1$ and $x=0$ and $x=\frac{3}{4}$
Work Step by Step
First derivative
\[
\begin{array}{l}
2 x(-1+x)^{2 / 3}+\frac{2}{3} x^{2}(-1+x)^{-1 / 3}= f^{\prime}(x) \\
=\frac{2 x^{2}+6 x(-1+x)}{3 \sqrt[3]{-1+x}}
\end{array}
\]
Points where the derivative is zero are stationary points.
\[
\frac{-6 x +8 x^{2}}{3 \sqrt[3]{-1+x}}=0
\]
A fraction is zero if the numerator is zero
\[
(-3+4 x)2 x=0
\]
Roots
\[
x= \frac{3}{4} \text { and } x=0
\]
Points where the tangent line is horizontal (thus, the derivative is zero) are critical points.
\[
x=\frac{3}{4} \text { and } x=1 \text { and } x=0
\]
Points where the first derivative is zero are stationary points.
\[
x=\frac{3}{4} \text { and } x=0
\]