Inverse functions

I'd start by testing a number of special cases and see what approach the trickier special cases suggest.

Simple special cases almost never suggest approaches to proof, but the tricky special cases often do.

 

Let [itex]h[/itex] be the function which takes [itex]x[/itex] to [itex]1/x[/itex]. Now if [itex]f(x)g(x) = 1[/itex] for all [itex]x[/itex] then [itex]f = h \circ g[/itex]. Then [itex]f^{-1} = g^{-1} \circ h^{-1}[/itex]. But [itex]h = h^{-1}[/itex] so [itex]f^{-1} = g^{-1} \circ h[/itex] as required.