#### Answer

$x=5$

#### Work Step by Step

$\dfrac{2}{x-3}-\dfrac{4}{x+3}=\dfrac{8}{x^{2}-9}$
Factor the denominator of the fraction on the right side of the equation:
$\dfrac{2}{x-3}-\dfrac{4}{x+3}=\dfrac{8}{(x-3)(x+3)}$
Multiply the whole equation by $(x-3)(x+3)$
$(x-3)(x+3)\Big[\dfrac{2}{x-3}-\dfrac{4}{x+3}=\dfrac{8}{(x-3)(x+3)}\Big]$
$2(x+3)-4(x-3)=8$
$2x+6-4x+12=8$
Take the $8$ to the left side of the equation and simplify the equation by combining like terms:
$2x+6-4x+12-8=0$
$-2x+10=0$
Solve for $x$:
$-2x=-10$
$x=\dfrac{-10}{-2}$
$x=5$
The initial equation is not undefine for the value of $x$ found. So the solution to this equation is $x=5$