Answer
f is increasing in $(-\infty, 3)$ and f is decreasing in $(3,+\infty)$
Work Step by Step
The function f is undefined when x=3, so 3 is not in the domain of f.
$f'(x)=\frac{16}{(9-3x)^{2}}$
$f'(x)>0$ so we have: f is increasing in $(-\infty, 3)$ and f is decreasing in $(3,+\infty)$