#### Answer

$\frac{9}{4(x-1)}$

#### Work Step by Step

$\frac{9}{x^{2}-1}$ $\div$ $\frac{12}{3x+3}$=
$\frac{9}{(x+1)(x-1)}$ $\times$ $\frac{3x+3}{12}$=
$\frac{9}{(x+1)(x-1)}$ $\times$ $\frac{3(x+1)}{12}$=
$\frac{9*3(x+1)}{(x+1)(x-1)12}$ =$\frac{27}{12(x-1)}$=
$\frac{9}{4(x-1)}$