Solve
\[\frac{|x^2 - 81|}{x^2 - 36x} < 0.\]
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
\[x \in \boxed{(0,9) \cup (9,36)}.\]