#### Answer

$\dfrac{3x}{x+3}$

#### Work Step by Step

$\dfrac{3^{-1}-x^{-1}}{9^{-1}-x^{-2}}=\dfrac{\dfrac{1}{3}-\dfrac{1}{x}}{\dfrac{1}{9}-\dfrac{1}{x^2}}=\dfrac{\dfrac{x-3}{3x}}{\dfrac{x^2-9}{9x^2}}=\dfrac{\dfrac{x-3}{3x}}{\dfrac{x^2-3^2}{9x^2}}=\dfrac{9x^2(x-3)}{3x(x^2-3^2)}=\dfrac{9x^2(x-3)}{3x(x+3)(x-3)}=\dfrac{3x}{x+3}$