Answer
$f^{-1}(x)=\frac{1}{x}+1$
Work Step by Step
$f(x)=\frac{1}{x-1},$
$y=\frac{1}{x-1},$
$x=\frac{1}{y-1},$
$y-1=\frac{1}{x},$
$y=\frac{1}{x}+1=f^{-1}(x),$
$(f^{-1}\circ f)(x)=\frac{1}{\frac{1}{x-1}}+1=\frac{x(x-1)}{x-1}=x$
$(f\circ f^{-1})(x)=\frac{1}{\frac{1}{x}+1-1}=x$
