Answer
$=\displaystyle \frac{3}{x-3}, \quad x\neq-4,3$
Work Step by Step
The fractions have a common denominator, we subtract the numerators...
$\displaystyle \frac{x^{2}+3x}{x^{2}+x-12}-\frac{x^{2}-12}{x^{2}+x-12}=\frac{x^{2}+3x-(x^{2}-12)}{x^{2}+x-12}$
$=\displaystyle \frac{3x+12}{x^{2}+x-12}$
factor what we can; for the denominator,
we search for two factors of c=-12 whose sum is b=1
$=\displaystyle \frac{3(x+4)}{(x+4)(x-3)}$
... exclude values that yield 0 in the denominator:
$x\neq-4,3$
... and, cancel the common factor
$=\displaystyle \frac{3}{x-3}, \quad x\neq-4,3$
