Answer
$ -\displaystyle \frac{x-14}{7},\quad x\neq-2,2.$
Work Step by Step
Start by recognizing a difference of squares in the denominator of the the denominator.
$\displaystyle \frac{\frac{3}{x-2}-\frac{4}{x+2}}{\frac{7}{x^{2}-4}}=\frac{\frac{3}{x-2}-\frac{4}{x+2}}{\frac{7}{(x-2)(x+2)}}$
Complex rational expressions have rational expressions in the numerator or/and in the denominator.
Here, multiplying both the numerator and denominator with $(x-2)(x+2)$ would get rid of these fractions.
Currently, we note that we have to exclude $\pm 2$ as they yield zero in corresponding denominators..
The denominator of the complex rational expression, $\displaystyle \frac{7}{(x-2)(x+2)}$
can not be zero as it has a nonzero numerator. No new exclusions.
$\displaystyle \frac{\frac{3}{x-2}-\frac{4}{x+2}}{\frac{7}{(x-2)(x+2)}}\times\frac{(x-2)(x+2)}{(x-2)(x+2)}=\frac{3(x+2)-4(x-2)}{7}$
$=\displaystyle \frac{3x+6-4x+8}{7}$
$=\displaystyle \frac{-x+14}{7}$
$=-\displaystyle \frac{x-14}{7},\quad x\neq-2,2.$
