Answer
$f^{-1}(x)=\frac{x+1}{x}$
$f(x)$: $\{x|x\ne1 \}$ and $\{y|y\ne0\}$.
$f^{-1}(x)$: $\{x|x\ne0\}$ and $\{y|y\ne1\}$.
Work Step by Step
1. $f(x)=\frac{1}{x-1} \Longrightarrow y=\frac{1}{x-1} \Longrightarrow x=\frac{1}{y-1} \Longrightarrow y=\frac{x+1}{x} \Longrightarrow f^{-1}(x)=\frac{x+1}{x}$
2. check $f(f^{-1}(x))=\frac{1}{\frac{x+1}{x}-1}=x$. $f^{-1}(f(x))=\frac{\frac{1}{x-1}+1}{\frac{1}{x-1}}=x$.
3. The domain and the range of $f(x)$: $\{x|x\ne1 \}$ and $\{y|y\ne0\}$.
The domain and the range of $f^{-1}(x)$: $\{x|x\ne0\}$ and $\{y|y\ne1\}$.
