if $f\left(x\right)$ is continuous and increasing function such that  the domain of $g\left(x\right)=\sqrt{f\left(x\right)-x}$ be R and $h\left(x\right)=\frac{1}{1-x}$ then the domain of $\varnothing \left(x\right)=\sqrt{f\left(f\right(f\left(x\right)\left)\right)-h\left(h\right(h\left(x\right)\left)\right)}$ is

$\begin{array}{l}h\left(x\right)=\frac{1}{1-x},x\ne 1\\ h\left(h\left(x\right)\right)=\frac{x-1}{x},x\ne 0,1\\ h\left(h(h\left(x\right)\right)=x,x\ne 0,1\\ andg\left(x\right)\ge 0\forall x\in R\\ \Rightarrow f\left(x\right)\ge x\\ \Rightarrow f\left(f\left(x\right)\right)\ge f\left(x\right)\ge x\\ (\therefore f\left(x\right)isincrese)\\ \Rightarrow f\left(f\left(f\left(x\right)\right)\right)\ge f\left(x\right)\ge x\\ \Rightarrow f\left(f\left(f\left(x\right)\right)\right)-x\ge 0\\ \forall x\in R-\{0,1\}\\ \therefore \varnothing \left(x\right)isdefinedforallx\in R-\{0,1\}\end{array}$