Solve the equation $\frac{1}{x^2+11x-8} + \frac{1}{x^2+2x-8} + \frac{1}{x^2-13x-8} = 0$

Since you require the sum to be zero, all you need is to compute the numerator of the LHS when put over a common denominator, since regardless of the value of the denominator--as long as it is nonzero--the LHS is zero only if the numerator is zero. Then once you solve for the roots of the numerator, check the validity of the solution set by substitution.

You started very well. To make things easier, set $A=x^2+7x-8$ (*) and the equation rewrites $$\frac{1}{A+4x} + \frac{1}{A-5x} + \frac{1}{A-20x} = 0.$$

Denominators are not allowed to be zeros. We solve $$(A-5x)(A-20x)+(A+4x)(A-20x)+(A+4x)(A-5x)=0$$ or equivalently $$3A^2-42Ax=0,$$ or even $$3(x^2+7x-8)(x^2-7x-8)=0,$$ which is easy to finish.

(*) I noticed that $x=1$ satisfies, and decided to make profit from it.