Решение:

$$\left(\frac{x+3}{x-3} + \frac{x-3}{x+3}\right) : \frac{15x}{x^2-9} = \left(\frac{(x+3)^2 + (x-3)^2}{(x-3)(x+3)}\right) : \frac{15x}{x^2-9} = \frac{x^2+6x+9 + x^2-6x+9}{x^2-9} : \frac{15x}{x^2-9} = \frac{2x^2+18}{x^2-9} : \frac{15x}{x^2-9} = \frac{2(x^2+9)}{x^2-9} \cdot \frac{x^2-9}{15x} = \frac{2(x^2+9)}{15x}$$