Answer
The inverse function is given by
$$f^{-1}(x)=\left(\frac{1-x}{1+x}\right)^2.$$
Work Step by Step
We have to express $x$ in terms of $y$. Then $x=f^{-1}(y)$.
$$y=\frac{1-\sqrt{x}}{1+\sqrt{x}}\Rightarrow(1+\sqrt{x})y=1-\sqrt{x}\Rightarrow y+y\sqrt{x}=1-\sqrt{x}.$$
Rearranging terms
$$y\sqrt{x}+\sqrt{x}=1-y\Rightarrow \sqrt{x}(y+1)=1-y\Rightarrow \sqrt{x}=\frac{1-y}{y+1},$$
which finally gives
$$x=\left(\frac{1-y}{1+y}\right)^2$$
so
$$f^{-1}(y)=\left(\frac{1-y}{1+y}\right)^2.$$
Renaming $y$ back to $x$ we have
$$f^{-1}(x)=\left(\frac{1-x}{1+x}\right)^2.$$
