Answer
(a)
Increasing on $(-\displaystyle \infty, \frac{-2}{\sqrt{7}})\cup(\frac{2}{\sqrt{7}}, \infty)$
Decreasing on $(\displaystyle \frac{-2}{\sqrt{7}},0)\cup(0,\frac{2}{\sqrt{7}})$
(b)
No absolute maximum.
Local maximum: $(-\displaystyle \frac{2}{\sqrt{7}}, 3.123)$
No absolute minimum.
Local minimum: $(\displaystyle \frac{2}{\sqrt{7}},-3.123).$
Work Step by Step
$h(x)=x^{7/3}-4x^{1/3}$
$h$ is defined everywhere
$h'(x)=\displaystyle \frac{7}{3}x^{4/3}- \displaystyle \frac{4}{3}x^{-2/3}=\frac{1}{3x^{2/3}}(7x^{2}-4)$
$h'(x)=\displaystyle \frac{(x\sqrt{7}+2)(x\sqrt{7}-2)}{3x^{2/3}}$
$h'$ is not defined at $ x=0\qquad$ ... critical point.
$h'(x)=0$ for $x=\displaystyle \frac{\pm 2}{\sqrt{7}}\approx\pm 0.756$ ... critical points.
$ h(\displaystyle \frac{-2}{\sqrt{7}})\approx 3.123 \qquad h(0)=0, \displaystyle \qquad h(\frac{-2}{\sqrt{7}})\approx-3.123$
Using testpoints in the intervals between critical points,
$h'(-8)=37 \gt 0$
$h'(-0.5)=-1.1905... \lt 0$
$h'(0.5)=-1.1905... \lt 0$
$h'(8)=37 \gt 0$
Tabular view:
$\begin{array}{l}
h':\\
\\
\\
h:
\end{array} \begin{array}{ccccccccc}
-\infty & & -\frac{2}{\sqrt{7}} & & 0 & & \frac{2}{\sqrt{7}} & & \infty\\
( & ++ & | & -- & )( & -- & & ++ & )\\
& & & & & & & & \\
& \nearrow & 3.123 & \searrow & 0 & \searrow & & \nearrow & \\
& & & & & & -3.123 & & \\
& & & & & & & &
\end{array} $
