$f(x)$ is discontinuous at $x=1$
Now $f(f(x))=\large\frac{1}{1-\Large\frac{1}{1-x}}$
$\Rightarrow \large\frac{x-1}{x}$ which is discontinuous at $x=0$
But $f(f(f(x)))=\large\frac{1}{1-\Large\frac{x-1}{x}}$$=x$
Which has no point of discontinuity.
Hence $x=0,1$ are the only points of discontinuity.
Hence (b) is the correct answer.