Clearly, $x=1$ is a point of discontinuity of the function $f\left(x\right)=\frac{x}{x-1}$ .
If $x\ne 1$ , then
$\left(fof\right)\left(x\right)=f\left(f\left(x\right)\right)=f\left(\frac{x}{x-1}\right)=x$ , which is continuous $\forall x\in R$ .
If $x\ne 1$ , then $\left(fofof\right)\left(x\right)=f\left(x\right)=\frac{x}{x-1}$
Hence, $f^{15}\left(x\right)={\left(fofof\dots \dots \dots \right)}^{15}\left(x\right)=f\left(x\right)$ which is continuous everywhere except $x=1$
So, the only point of discontinuity is $x=1$