Answer
-$2$
Work Step by Step
$f(x)$ = $\frac{x+6}{x-2}$, $x$ $\gt$ $2$, $a$ = $3$
$\frac{x+6}{x-2}$ = $3$
$x+6$ = $3(x-2)$
$x+6$ = $3x-6$
$x$ = $6$
$f(6)$ = $3$ $so$ $f^{-1}(3)$ = $6$
$(f^{-1})$'$(3)$ = $\frac{1}{f'(f^{-1}(3))}$ = $\frac{1}{f'(6)}$
$f'(x)$ = $\frac{(x-2)\frac{d(x+6)}{dx}-(x+6)\frac{d(x-2)}{dx}}{(x-2)^{2}}$ = $\frac{(x-2)(1)-(x+6)(1)}{(x-2)^{2}}$ = $\frac{x-2-x-6}{(x-2)^{2}}$ = $\frac{-8}{(x-2)^{2}}$
$f'(6)$ = $\frac{-8}{(6-2)^{2}}$ = -$\frac{8}{16}$ = -$\frac{1}{2}$
$so$ $(f^{-1})$'$(3)$ = $\frac{1}{f'(6)}$ = $\frac{1}{\frac{-1}{2}}$ = -$2$
