\(\left(\frac{9}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)\)
\(=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)
\(=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\frac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\frac{3x\left(9+x^2-3x\right)\left(x+3\right)}{x\left(x-3\right)\left(x+3\right)\left(3x-9-x^2\right)}=\frac{3\left(9+x^2-3x\right)}{-\left(x-3\right)\left(x^2-3x+9\right)}=-\frac{3}{x-3}\)