Answer
(a)
Increasing on $(-1,0)\cup(1, \infty)$
Decreasing on $(-\infty, -1)\cup(0,1)$
(b)
No absolute maximum.
Local maximum: $(0,0)$
Absolute minimum: $(-1,-3)$ and $(1,-3)$.
No other local minima.
Work Step by Step
$k(x)=x^{8/3}-4x^{2/3}$
$k$ is defined everywhere
$k'(x)=\displaystyle \frac{8}{3}x^{5/3}- \displaystyle \frac{8}{3}x^{-1/3}=\frac{8}{3x^{1/3}}(x^{2}-1)$
$k'(x)=\displaystyle \frac{8(x+1)(x-1)}{3x^{1/3}}$
$k'$ is not defined at $ x=0\qquad$ ... critical point.
$k'(x)=0$ for $x=\pm 1$... critical points.
$ k(-1)-3 \qquad k(0)=0, \qquad k(1)=-3$
Using testpoints in the intervals between critical points,
$k'(-8)=-84 \lt 0$
$k'(-0.5)=2.5198... \gt 0$
$k'(0.5)=-2.5198.... \lt 0$
$k'(8)=84 \gt 0$
Tabular view:
$\begin{array}{l}
h':\\
\\
\\
h:
\end{array} \begin{array}{lllllllll}
-\infty & & -1 & & 0 & & 1 & & \infty\\
( & -- & | & ++ & )( & -- & | & ++ & )\\
& & & & & & & & \\
& \searrow & & \nearrow & 0 & \searrow & & \nearrow & \\
& & -3 & & & & -3 & & \\
& & & & & & & &
\end{array} $
