Answer
$y^{-1}=\ln\left(\frac{y+1}{1-y}\right)$
Work Step by Step
To find the inverse of the function, solve the equation for $x$:
$$y=\frac{1-e^{-x}}{1+e^{-x}}$$
$$y(1+e^{-x})=1-e^{-x}$$
$$y+ye^{-x}=1-e^{-x}$$
$$y-1=-ye^{-x}-e^{-x}$$
$$y-1=-e^{-x}(y+1)$$
$$\frac{y-1}{y+1}=-e^{-x}$$
$$\frac{1-y}{y+1}=e^{-x}$$
$$\ln\left(\frac{1-y}{y+1}\right)=-x$$
$$-\ln\left(\frac{1-y}{y+1}\right)=x$$
$$\ln\left(\frac{y+1}{1-y}\right)=x$$
so the inverse is:
$$y^{-1}=\ln\left(\frac{y+1}{1-y}\right)$$
