Find the domain and the correspondence rule of $F$ $F=f ∘ f∘ f$, $f(x)=\frac{1}{1-x} ⇒ Dom f(x)=\mathbb{R}-\{1\} \land Ran f(x)=\mathbb{R}-\{0\}$, then:
\begin{align*}                                                                                                                                                  
 Dom(f∘ f)&amp;=\{ x ∈ Dom(f) \land f(x) ∈ Dom(f)\} \\                                                                                                              
           &amp;=\{ x ∈ \mathbb{R}-\{1\} \land \frac{1}{1-x} ∈ \mathbb{R}-\{1\} \} \\                                                                               
           &amp;=\{ x ∈ \mathbb{R}-\{1\} \land x ∈ \mathbb{R}-\{0\}-\{1\}  \} \\                                                                                    
                &amp;=\{ x ∈ \mathbb{R}-\{1\}  \} \\                                                                                                                
  Dom(f∘ f∘ f)&amp;=\{ x ∈ Dom(f∘ f) \land f(f(x)) ∈ Dom(f∘ f)\} \\                                                                                                 
          &amp;=\{ x ∈  x ∈ \mathbb{R}-\{1\} \land \frac{1}{1-\dfrac{1}{1-x}} x ∈ \mathbb{R}-\{1\}\} \\                                                             
            &amp;=\{ x ∈ \mathbb{R}-\{1\} \land 1-x ∈                                                                                                               
              \mathbb{R}-\{1\}\} \\                                                                                                                             
  &amp;=\{ x ∈ \mathbb{R}-\{1\} \} \\                                                                                                                              
F&amp;=x
\end{align*}
It's wrong?

From: Lumbreras Editors
Since $f(\cdot)$ only takes values $\neq 1$, $f(f(x))$ is defined for $x \neq 1$ (because we need that the inner $f(x)$ is defined) and all the values of $x$ for which $f(x) \neq 1$, that is, $f(f(x))$ is defined for $x \neq 0,1$. Similarly, $f(f(f(x)))$ is defined for $x \neq 0,1$ (because we need that $f(f(x))$ is defined) and all the values of $x$ for which $f(f(x)) \neq 1$, that is, $f(f(f(x)))$ is defined for $x \neq 0,1$ since $f(f(x))$ is always $\neq 1$.
Thus $\operatorname{dom}(F) = \mathbb R \setminus \{0,1\}$.
Since $f(\cdot)$ only takes values $\neq 1$, $f(f(x))$ is defined for $x \neq 1$ (because we need that the inner $f(x)$ is defined) and all the values of $x$ for which $f(x) \neq 1$, that is, $f(f(x))$ is defined for $x \neq 0,1$. Similarly, $f(f(f(x)))$ is defined for $x \neq 0,1$ (because we need that $f(f(x))$ is defined) and all the values of $x$ for which $f(f(x)) \neq 1$, that is, $f(f(f(x)))$ is defined for $x \neq 0,1$ since $f(f(x))$ is always $\neq 1$.
Thus $\operatorname{dom}(F) = \mathbb R \setminus \{0,1\}$.