All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
$f^{2}(x)$
$=f[f(x)]$
$=f(\frac{1}{1-x})$
$=\frac{1}{1-\frac{1}{1-x}}$
$=\frac{1}{\frac{1-x-1}{1-x}}$
$=\frac{x-1}{x}. x\neq{0}, x\neq{1}$

let $y=f(x)$
$y(1-x)=1$
$x=\frac{y-1}{y}$
$f^{-1}(x)=\frac{x-1}{x}, x\neq{1}, x\neq{0}$

Since $D_{f^{-1}}=R_{f}=\mathbb{R} \backslash \{0,1\}$

We have that $f^{2}(x)=f^{-1}(x)$

(ii)
$f^{2}(x)=f^{-1}(x)$
$f[f^{2}(x)]=f[f^{-1}(x)]$
$f^{3}(x)=x, x\neq{1}, x\neq{0}$

Students must understand what it means for two functions to be equal. It isn’t just about the rule (loosely put, formula) the same, but they require the same domain, and as a result the same range too. Stating the domains in (i) is important to show true understanding of what equal functions mean. It is saddening that some students also thought that $f^{2}(x)=f(x)f(x)$ which is not the case. $f^{2}(x)$ is a composite function, “f in f” while $f(x)f(x)$ is a product of two functions.