Clearly, $x=1$ is a point of discontinuity of the function $f\left(x\right)=\frac{1}{1-x}$.
if $x\ne 1$, then
$\left(fof\right)\hspace{0.17em}\left(x\right)=f\left[f\left(x\right)\right]=f\left(\frac{1}{1-x}\right)=\frac{x-1}{x},$ which is discontinuous at $x=0$.
If $x\ne 0$ and $x\ne 1$, then
$\left(fo\hspace{0.17em}fof\right)\left(x\right)=f\left[\left(fof\right)\left(x\right)\right]=f\left(\frac{x-1}{x}\right)=x$
Which is continuous everywhere.
Hence, $f^{30}\left(x\right)=x,$ which is continuous everywhere.
So, the only points of discontinuity are $x=0$ and $x=1$