Answer
(a)
Increasing on $(-2, 0)\cup(0, \infty)$
Decreasing on $(-\infty, -2)$
(b)
No absolute maximum.
No local maximum.
Absolute minimum: $(-2, -6\sqrt[3]{2})$.
No other local minimum.
Work Step by Step
$f(x)=x^{4/3}+8x^{1/3}$
$f$ is defined everywhere
$f'(x)=\displaystyle \frac{4}{3}x^{1/3}+ \displaystyle \frac{8}{3}x^{-2/3}=\frac{4}{3x^{2/3}}(x+2)$
$f'$ is not defined at $ x=0\qquad$ ... critical point.
$f'(x)=0$ for $x=-2$ ... critical point.
$ f(-2)=-6\sqrt[3]{2}\approx-7.56\qquad f(0)=0.$
Using testpoints in the intervals between critical points,
$f'(-8)=-2\lt 0$
$f'(-1)=1.333... \gt 0$
$f'(8)=3.333... \gt 0$
Tabular view:
$\begin{array}{l}
f':\\
\\
\\
f:
\end{array} \begin{array}{lllllll}
-\infty & & -2 & & 0 & & \infty\\
( & -- & | & ++ & )( & ++ & )\\
& & & & & & \\
& \searrow & & \nearrow & 0 & \nearrow & \\
& & -6\sqrt[3]{2} & & & & \\
& & & & & &
\end{array} $
