Find the sum of all $x$ that satisfy the equation $\frac{-9x}{x^2-1} = \frac{2x}{x+1} - \frac{6}{x-1}.$
Let's look at the right hand side: $\frac{2x(x-1)-6(x+1)}{x^2-1}=\frac{2x^2-2x-6x-6}{x^2-1}=\frac{2x^2-8x-6}{x^2-1}$. So $2x^2-8x-6=-9x\implies 2x^2+x-6=0$, so the sum of the solutions is $\boxed{-\frac12}$ by Vieta's formulae.