Упростим выражение:
$$\frac{3}{x+3} + \frac{2x}{x^2-3x} + \frac{3}{9-x^2} = \frac{3}{x+3} + \frac{2x}{x(x-3)} - \frac{3}{x^2-9} = \frac{3}{x+3} + \frac{2}{x-3} - \frac{3}{(x-3)(x+3)}$$
$$\frac{3(x-3) + 2(x+3) - 3}{(x-3)(x+3)} = \frac{3x - 9 + 2x + 6 - 3}{(x-3)(x+3)} = \frac{5x - 6}{(x-3)(x+3)} = \frac{5x-6}{x^2-9}$$