Bài giải:

$\frac{18}{(x-3)({{x}^{2}}-9)}-\frac{3}{{{x}^{2}}-6x+9}-\frac{x}{{{x}^{2}}-9}$= $\frac{18}{(x-3)(x-3)(x+3)}-\frac{3}{{{(x-3)}^{2}}}-\frac{x}{(x-3)(x+3)}$

= $\frac{18-3(x+3)-x(x-3)}{(x-3)(x-3)(x+3)}$ = $\frac{18-3x-9-{{x}^{2}}+3x}{(x-3)(x-3)(x+3)}$ = $\frac{9-{{x}^{2}}}{(x-3)(x-3)(x+3)}$

= $\frac{-(x-3)(x+3)}{(x-3)(x-3)(x+3)}$ = $\frac{-1}{x-3}$

Vậy đáp án đúng là: D