jmc

Note that $|x^2-81|\ge 0$ for all $x,$ with $|x^2-81|=0$ only for $x=\pm 9.$

The denominator factors as $x(x-36).$ This is negative only for $0<x<36.$ Thus, the solution is $$x\in \boxed{(0,9)\cup (9,36)}.$$