Complete solution set of $|\frac{x^2}{x-1}|\le 1$ is given by:

$|\frac{x^2}{x-1}|\le 1⟹\frac{x^2}{|x-1|}\le 1$ (as $x^2\ge 0$).$x^2\le |x-1|$ where $x\ne 1$.Case 1: $x>1$$x^2\le x-1⟹x^2-x+1\le 0$.The discriminant $D=(-1{)}^2-4(1)(1)=-3<0$.Since the leading coefficient is positive and $D<0$, $x^2-x+1$ is always positive.So, no solution for $x>1$.Case 2: $x<1$$x^2\le -(x-1)⟹x^2\le 1-x⟹x^2+x-1\le 0$.Roots of $x^2+x-1=0$ are $x=\frac{-1\pm \sqrt{1+4}}{2}=\frac{-1\pm \sqrt{5}}{2}$.The inequality holds between the roots: $\frac{-1-\sqrt{5}}{2}\le x\le \frac{-1+\sqrt{5}}{2}$.Since $\frac{-1+\sqrt{5}}{2}\approx 0.618<1$, all these values satisfy $x<1$.Thus, the solution is $[\frac{-1-\sqrt{5}}{2},\frac{-1+\sqrt{5}}{2}]$.