We should confirm that f(x) has an inverse in the first place. If it does, then
[tex]f\left(f^{-1}(x)\right) = x[/tex]
Given that f(x) = x/(2 - x), we have
[tex]f\left(f^{-1}(x)\right) = \dfrac{f^{-1}(x)}{2 - f^{-1}(x)} = x[/tex]
Solve for the inverse:
[tex]\dfrac{f^{-1}(x)}{2 - f^{-1}(x)} = x[/tex]
[tex]f^{-1}(x) = 2x - x f^{-1}(x)[/tex]
[tex]f^{-1}(x) + x f^{-1}(x) = 2x[/tex]
[tex]f^{-1}(x) (1 + x) = 2x[/tex]
[tex]f^{-1}(x) = \dfrac{2x}{1+x}[/tex]
Then
[tex]f^{-1}(-2) = \dfrac{2(-2)}{1-2} = \boxed{4}[/tex]
Note that this is the same as solve for x when f(x) = -2 :
x/(2 - x) = -2 ⇒ x = 4
