$$ \frac{3}{x-3} - \frac{x+15}{x^2-9} - \frac{2}{x} = \frac{3}{x-3} - \frac{x+15}{(x-3)(x+3)} - \frac{2}{x} $$
Приведем дроби к общему знаменателю $$x(x-3)(x+3)$$:
$$ \frac{3x(x+3) - (x+15)x - 2(x-3)(x+3)}{x(x-3)(x+3)} = \frac{3x^2 + 9x - x^2 - 15x - 2(x^2-9)}{x(x-3)(x+3)} = \frac{3x^2 + 9x - x^2 - 15x - 2x^2 + 18}{x(x-3)(x+3)} = \frac{-6x + 18}{x(x-3)(x+3)} = \frac{-6(x-3)}{x(x-3)(x+3)} = \frac{-6}{x(x+3)} $$
Ответ: $$\frac{-6}{x(x+3)}$$