Let $z=x+iy,$ where $x,y\in\mathbb{R}$ and $z\ne 1$.

Given that $\frac{z^2}{z-1}$ is real. A complex number is real iff its imaginary part is zero.

Compute: $z^2=(x+iy)^2=x^2-y^2+2ixy,$ and $z-1=(x-1)+iy.$

So, $\frac{z^2}{z-1}=\frac{(x^2-y^2+2ixy)}{(x-1)+iy}.$

Rationalize the denominator: $\frac{z^2}{z-1}=\frac{(x^2-y^2+2ixy)\big((x-1)-iy\big)}{(x-1)^2+y^2}.$

We only need the imaginary part of the numerator.

Let $A=x^2-y^2,\qquad B=2xy.$ Then numerator is $(A+iB)\big((x-1)-iy\big).$

Its imaginary part is $B(x-1)-Ay.$

Therefore, $2xy(x-1)-(x^2-y^2)y=0.$

Simplify: $2x^2y-2xy-x^2y+y^3=0$ $x^2y-2xy+y^3=0$ $y(x^2-2x+y^2)=0.$

Hence either $y=0$ or $x^2-2x+y^2=0.$

Interpret geometrically:

• $y=0$ represents the real axis.
• $x^2-2x+y^2=0$ $x^2-2x+1+y^2=1$ $(x-1)^2+y^2=1,$ which is a circle with centre $(1,0)$ and radius $1$.

Check whether this circle passes through the origin: At $(0,0)$, $(0-1)^2+0^2=1,$ so the origin lies on the circle.

Therefore, the point represented by $z$ lies either on the real axis or on a circle passing through the origin.

Hence the correct option is: $\boxed{\text{A}}$