Answer
$-\displaystyle \frac{9x^{3}}{(x-3)^{2}}$
Work Step by Step
$...=\displaystyle \frac{3+x}{3-x}\div\frac{x^{2}-9}{9x^{3}}=\frac{3+x}{3-x}\cdot\frac{9x^{3}}{x^{2}-9}$
Factoring,
$ x^{2}-9=(x+3)(x-3)\quad$ (a difference of squares)
$3-x=-(x-3)$
... = $\displaystyle \frac{(x+3)\cdot 9x^{3}}{-(x-3)(x-3)(x+3)}$
... cancel the following from both sides of the fraction line:
$(x+3)$
... = $\displaystyle \frac{9x^{3}}{-(x-3)(x-3)}$
= $-\displaystyle \frac{9x^{3}}{(x-3)^{2}}$
