Let $f$ be a composite function of $x$ defined by $f\left(u\right)=\frac{1}{u^2+u-2}, u\left(x\right)=\frac{1}{x-1}.$ Then the number of points $x$ where $f$ is discontinuous is

$u\left(x\right)=\frac{1}{x-1},$ which is discontinuous at $x=1$

$f\left(u\right)=\frac{1}{(u+2)(u-1)},$ which is discontinuous at $u=-2,1$

When $u=-2,$ then $\frac{1}{x-1}=-2\Rightarrow x=\frac{1}{2}$

When $u=-1,$ then $\frac{1}{x-1}=1\Rightarrow x-1=1\Rightarrow x=2$

Discontinuous at $x=1,\frac{1}{2},2$