The equation $y(x-1)=x^2-1$ can be graphed on Cartesian plane by inserting values in for $y$ and then solving for $x$
For example, if $y=3$ then:
$3(x-1)=x^2-1$
$3x-3=x^2-1$
$3x=x^2+2$
$0=x^2-3x+2$
$0=(x-2)(x+1)$
Thus $x=1$ or $2$ and so co-ordinates (1,3) and (2,3) can be plotted.
Repeating this process with different values for $y$ produces this graph with a vertical line.
Is the vertical line that appears in the graph a removable singularity or something else?
Let’s rewrite your equation: $$x^2-1-y(x-1) = (x-1)(x-y+1) = 0,$$ so it’s the product of the equations $x-1=0$ and $x-y+1=0$. The graph of your original implicit equation is therefore the union of the graphs of these two equations, i.e., it’s a pair of lines. One of them happens to be vertical, but that doesn’t mean that there’s a singularity per se. It simply means that $y$ can’t be expressed as a function of $x$ everywhere on the plane.