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[tex]\frac{{3-3x}}{{x^2-9}}+\frac{x}{{x-3}}+\frac{2}{{x+3}}-\frac{2}{3}=\frac{-9x+9}{3x^2-27}+\frac{3x^2+9x}{3x^2-27}+\frac{6x-18}{3x^2-27}-\frac{2x^2-18}{3x^2-27}=\\ \frac{-9x+9+3x^2+9x+6x-18-2x^2+18}{3x^2-27}=\fra{x^2+6x+9}{3x^2-27}=\frac{(x+3)(x+3)}{3(x-3)(x+3)}=\\ \frac{x+3}{3(x-3)}[/tex]