$\dfrac{x}{x-3} -\dfrac{2}{x+3} +\dfrac{x(1-x)}{x^2 - 9}\qquad (x\ne \pm 3)$
$=\dfrac{x(x+3)}{(x-3)(x+3)} -\dfrac{2(x-3)}{(x-3)(x+3)} +\dfrac{x -x^2}{(x-3)(x+3)}$
$=\dfrac{x^2 + 3x - 2x + 6 + x - x^2}{(x-3)(x+3)}$
$=\dfrac{2x +6}{(x-3)(x+3)}$
$=\dfrac{2(x+3)}{(x-3)(x+3)}$
$=\dfrac{2}{x-3}$