Answer
$=\displaystyle \frac{(x-3)(x+3)}{x(x+4)},\qquad x\neq 0,-4,3$
Work Step by Step
Factor what you can:
$x^{2}-9=$ difference of squares $= (x-3)(x+3)$
$x^{2}$ is factored
$x^{2}-3x=x(x-3)$
Factoring $x^{2}+bx+c$ = $x^{2}+x-12 $
we search for two factors of c (m and n) such that m+n=b.
If they exist, $x^{2}+bx+c =(x+m)(x+n)$
$x^{2}+x-12 = \quad$... we find factors $-3$ and $+4,$
$=(x+4)(x-3)$
Expression $=\displaystyle \frac{(x-3)(x+3)}{x^{2}}\cdot\frac{x(x-3)}{(x+4)(x-3)}\qquad$
...exclude the values that yield 0 in the denominator:
$x\neq 0,-4,3$
... cancel common factors: $x$, and $(x-3)$
$=\displaystyle \frac{(x-3)(x+3)}{x(x+4)},\qquad x\neq 0,-4,3$
