If $z\ne 1, \frac{z^2}{z-1}$ is real, then point represented by the complex number $z$ lies 

As $\frac{z^2}{z-1}$ is real, we get 

$\begin{array}{l} \frac{z^2}{z-1}=\frac{{\overline{z}}^2}{\overline{z}-1}\\ \iff z^2(\overline{z}-1)={\overline{z}}^2(z-1)\\ \iff z\overline{z}(z-\overline{z})-(z-\overline{z})(z+\overline{z})=0\\ \iff (z-\overline{z})(z\overline{z}-z-\overline{z})=0\\ \Rightarrow z-\overline{z}=0\text{ or }z\overline{z}-z-\overline{z}=0\end{array}$
$\Rightarrow$$z$ lies on the real axis or $z$ lies on a circle through the origin.