Answer
(a)
Increasing on $(-\infty, 1)\cup(3, \infty)$
Decreasing on $(1, 2)\cup(2, 3)$
(b)
No absolute maximum.
Local maximum: $(1,2)$
No absolute minimum.
Local minimum: $(3,6)$
Work Step by Step
$f$ is defined everywhere except $x=2.$
$f'(x)=\displaystyle \frac{2x(x-2)-(x^{2}-3)(1)}{(x-2)^{2}}=\frac{x^{2}-4x+3}{(x-2)^{2}}$
$f'(x)= \displaystyle \frac{(x-3)(x-1)}{(x-2)^{2}}$
Which is undefined at $ x=2$: not a critical point (not in the domain).
$f'(x)=0$ for $x=1,\ 3$: critical points.
$f(1)=2,\quad f(3)=6$
Using testpoints in the intervals between critical points,
$f'(0)=0.75 \gt 0$
$f'(1.5)=-3\lt 0$
$f'(2.5)=-3\lt 0$
$f'(4)=0.75 \gt 0$
Tabular view:
$\begin{array}{l}
f':\\
\\
\\
f:
\end{array} \begin{array}{cccccccc}
-\infty & & 1 & & 2 & & 3 & & \infty\\
( & ++ & & -- & )( & -- & | & ++ & )\\
& & & & & & & & \\
& \nearrow & 2 & \searrow & )( & \searrow & & \nearrow & \\
& & & & & & 6 & & \\
& & & & & & & &
\end{array} $
(a)
Increasing on $(-\infty, 1)\cup(3, \infty)$
Decreasing on $(1, 2)\cup(2, 3)$
(b)
No absolute maximum.
Local maximum: $(1,2)$
No absolute minimum.
Local minimum: $(3,6)$
