Answer
$=\displaystyle \frac{x-2}{x+2}, \quad x\neq-2,3$
Work Step by Step
The fractions have a common denominator, we subtract the numerators...
$\displaystyle \frac{x^{2}-4x}{x^{2}-x-6}-\frac{x-6}{x^{2}-x-6}=\frac{x^{2}-4x-(x-6)}{x^{2}-x-6}$
$=\displaystyle \frac{x^{2}-5x+6}{x^{2}-x-6}$
... factor what we can; for trinomials $x^{2}+bx+c,$
we search for two factors of c whose sum is b
$=\displaystyle \frac{(x-2)(x-3)}{(x-3)(x+2)}$
... exclude values that yield 0 in the denominator:
$x\neq-2,3$
... and, cancel the common factor
$=\displaystyle \frac{x-2}{x+2}, \quad x\neq-2,3$
