Answer
(a)
Increasing on $(-\infty, 0)\cup(0, \infty)$
Not decreasing anywhere
(b)
No absolute maximum.
No local maximum.
No absolute minimum.
No local minimum.
Work Step by Step
$f$ is defined everywhere
$f'(x)=\displaystyle \frac{3x^{2}(3x^{2}+1)-x^{3}(6x)}{(3x^{2}+1)^{2}}=\frac{3x^{2}(x^{2}+1)}{(3x^{2}+1)^{2}}$
$f'$ defined everywhere$\qquad$
$f'(x)=0$ for $x=0$: critical point.
$f(0)=0$
Using testpoints in the intervals between critical points,
$f'(-1)=0.375 \gt 0$
$f'(1)=0.375 \gt 0$
Tabular view:
$\begin{array}{l}
f':\\
\\
\\
f:
\end{array} \begin{array}{lllll}
-\infty & & 1 & & \infty\\
( & ++ & | & ++ & )\\
& & & & \\
& \nearrow & 0 & \nearrow & \\
& & & & \\
& & & &
\end{array} $
(a)
Increasing on $(-\infty, 0)\cup(0, \infty)$
Not decreasing anywhere
(b)
No absolute maximum.
No local maximum.
No absolute minimum.
No local minimum.
