Answer
$$\frac{2(7x-20)}{(x+4)^2(x-4)}$$
Work Step by Step
Creating common denominators to simplify: $$\frac{2}{x^2-16}-\frac{3x}{x^2+8x+16}+\frac{3}{x+4}=\frac{14x-40}{(x+4)^2(x-4)} \\ \frac{2}{(x+4)(x-4)}-\frac{3x}{x^2+8x+16}+\frac{3}{x+4} \\ \frac{2(x+4)}{(x+4)^2(x-4)}-\frac{3x(x-4)}{(x+4)^2(x-4)}+\frac{3(x+4)(x-4)}{(x+4)^2(x-4)} \\ \frac{2(x+4)-3x(x-4)+3(x+4)(x-4)}{(x+4)^2(x-4)} \\ \frac{14x-40}{(x+4)^2(x-4)} \\ \frac{2(7x-20)}{(x+4)^2(x-4)}$$
