#### Answer

The maximum is $f(-1)=3$; the minimum is $f(1) = -1$

#### Work Step by Step

Given $$f(x)=x^{2 / 3}-2 x^{1 / 3}, \quad[-1,3]$$
Since
$$ f'(x)= \frac{2}{3x^{\frac{1}{3}}}-\frac{2}{3x^{\frac{2}{3}}}$$
We find the critical points
\begin{align*}
f'(x)&=0\\
\frac{2x^{\frac{1}{3}}-2}{3x^{\frac{2}{3}}}&=0\\
2x^{\frac{1}{3}}-2&=0
\end{align*}
Then $f(x)$ has critical points at $x= 1,\ x=0$.
We check the values at the critical points and end points:
\begin{align*}
f(-1)&= 3 \\
f(0)&=0 \\
f(1)&= -1 \\
f(3)&\approx-0.804
\end{align*}
Then the maximum is $f(-1)=3$ and the minimum is $f(1) = -1$