Consider a function $f(x)$ such as $x\mapsto 2e^x-\frac1{e^x}$. How do you find $f^{-1}(x)$?
I have tried, logarithms, squaring, substitution, but I wasn't able to isolate $x$. The correct answer, according to Wolfram Alpha is $f^{-1}(x) = \log{\left(\frac14\left(x+\sqrt{x^2+8}\right)\right)}$.
If you start with
$$x=2e^y-e^{-y}$$
and multiply both sides by $e^y$, you get
$$x e^y=2e^{2y}-1$$
which you can treat as a quadratic equation in $e^y$. Use the quadratic formula to solve for $e^y$ (be careful in choosing roots!), undo the exponential with the logarithm, and you have your needed result.