Let $f(x)=\frac{1}{1-x}$. Then, $\{f o(f o f)\}(x)$

Solution:
Domain of $f$ :
$1-x \neq 0$
$\Rightarrow x \neq 1$
Domain of $f=R-\{1\}$
Range of $f$ :
$y=\frac{1}{1-x}$
$\Rightarrow 1-x=\frac{1}{y}$
$\Rightarrow x=1-\frac{1}{y}$
$\Rightarrow x=\frac{y-1}{y}$
$\Rightarrow y \neq 0$
Range of $f=R-\{0\}$
So, $f: R-\{1\} \rightarrow R-\{0\}$ and $f: R-\{1\} \rightarrow R-\{0\}$
Range of $f$ is not a subset of the domain of $f$.
Domain ( of $f)=\{x: x \in$ domain of $f$ and $f(x) \in$ domain of $f\}$
Domain ( $f o f)=\left\{x: x \in R-\{1\}\right.$ and $\left.\frac{1}{1-x} \in R-\{1\}\right\}$
Domain ( $f o f)=\left\{x: x \neq 1\right.$ and $\left.\frac{1}{1-x} \neq 1\right\}$
Domain ( $f o f)=\{x: x \neq 1$ and $1-x \neq 1\}$
Domain ( $f o f)=\{x: x \neq 1$ and $x \neq 0\}$
Domain ( $f o f)=R-\{0,1\}$
$(f \circ f)(x)=f(f(x))=f\left(\frac{1}{1-x}\right)=\frac{1}{1-\frac{1}{x}}=\frac{1-x}{1-x-1}=\frac{1-x}{-x}=\frac{x-1}{x}$
For range of fof,$x \neq 0$
Now, $f o f: R-\{0,1\} \rightarrow R-\{0\}$ and $f: R-\{1\} \rightarrow R-\{0\}$
Range of $f o f$ is not a subset of domain of $f .$
Domain $(f o(f o f))=\{x: x \in$ domain of $f o f$ and $(f o f)(x) \in$ domain of $f\}$
Domain $(f o(f o f))=\left\{x: x \in R-\{0,1\}\right.$ and $\left.\frac{x-1}{x} \in R-\{1\}\right\}$
Domain $(f o(f o f))=\left\{x: x \neq 0,1\right.$ and $\left.\frac{x-1}{x} \neq 1\right\}$
Domain $(f o(f o f))=\{x: x \neq 0,1$ and $x-1 \neq x\}$
Domain $(f o(f o f))=\{x: x \neq 0,1$ and $x \in R\}$
Domain $(f o(f o f))=R-\{0,1\}$
$(f o(f o f))(x)=f((f o f)(x))$
$=f\left(\frac{x-1}{x}\right)$
$=\frac{1}{1-\frac{x-1}{x}}$
$=\frac{x}{x-x+1}$
$=x$
So, $(f o(f o f))(x)=x$, where $x \neq 0,1$
So, the answer is (c).