If $$f\left( x \right) = \frac{1}{{1 - x}},$$   then the points of discontinuity of the function $$f\left[ {f\left\{ {f\left( x \right)} \right\}} \right]$$   are :

Solution :
We have, $$f\left( x \right) = \frac{1}{{1 - x}}$$
As at $$x = 1,\,f\left( x \right)$$   is not defined, $$x = 1$$  is a point of discontinuity of $$f\left( x \right).$$
If $$x \ne 1,\,f\left[ {f\left( x \right)} \right] = f\left( {\frac{1}{{1 - x}}} \right) = \frac{1}{{1 - \frac{1}{{\left( {1 - x} \right)}}}} = \frac{{x - 1}}{x}$$
$$\therefore \,x = 0,\,1$$   are points of discontinuity of $$f\left[ {f\left( x \right)} \right].$$
If $$x \ne 0,\,x \ne 1$$
$$f\left[ {f\left\{ {f\left( x \right)} \right\}} \right] = f\left( {\frac{{x - 1}}{x}} \right) = \frac{1}{{1 - \frac{{\left( {x - 1} \right)}}{x}}} = x$$