Calculus: Early Transcendentals, Chapter 1, 1.6, Section 1.6, Problem 26

Determine a formula for the inverse function $\displaystyle y= \frac{e^x}{1 + 2e^x}$

To find the inverse of $f$, we must first write
$\displaystyle y = \frac{e^x}{1 + 2e^x}$
Then we solve this equation for $x$

$
\begin{equation}
\begin{aligned}
y ( 1 + 2e^x) &= e^x\\
\\
y + 2e^x y &= e^x\\
\\
2e^x y - e^x &= - y\\
\\
e^x(2y - 1) &= -y\\
\\
e^x &= \frac{-y}{2y - 1}\\
\\
\ln e^x &= \ln \left[ \frac{-y}{2y - 1} \right]\\
\\
x &= \ln \left[ \frac{-y}{2y - 1} \right]
\end{aligned}
\end{equation}
$


Finally, we interchange $x$ and $y$:
$\displaystyle y = \ln \left[ \frac{-x}{2x - 1} \right]$

Therefore, the inverse function is
$\displaystyle f^{-1}(x) = \ln \left[ \frac{-x}{2x - 1} \right]$