Answer
(a)
Increasing on $(-\infty, -2)\cup(0, \infty)$
Decreasing on $(-2, 0)$
(b)
No absolute maximum.
Local maximum: $(-2, 3\sqrt[3]{4})$
No absolute minimum.
Local minimum: $(0,0).$
Work Step by Step
$g(x)=x^{5/3}+5x^{2/3}$
$g$ is defined everywhere
$g'(x)=\displaystyle \frac{5}{3}x^{2/3}+ \displaystyle \frac{10}{3}x^{-1/3}=\frac{5}{3x^{1/3}}(x+2)$
$g'$ is not defined at $ x=0\qquad$ ... critical point.
$g'(x)=0$ for $x=-2$ ... critical point.
$ g(-2)=3\sqrt[3]{4}\approx 4.76 \qquad g(0)=0.$
Using testpoints in the intervals between critical points,
$g'(-8)=5 \gt 0$
$g'(-1)=-1.667... \lt 0$
$g'(8)=8.333... \gt 0$
Tabular view:
$\begin{array}{l}
f':\\
\\
\\
f:
\end{array} \begin{array}{lllllll}
-\infty & & -2 & & 0 & & \infty\\
( & ++ & | & -- & )( & ++ & )\\
& & & & & & \\
& \nearrow & 3\sqrt[3]{4} & \searrow & & \nearrow & \\
& & & & 0 & & \\
& & & & & &
\end{array}$
