Help finding inverse of $f(x)=\frac{e^x-e^{-x}}{2}$

You have $$ (e^y)^2 - 2x(e^y) - 1 = 0 $$ From this you get $$ e^y = \frac{2x\pm \sqrt{{\color{red}{4x}^2} + 4}}{2}= x\pm \sqrt{x^2 + 1} $$ Here you only get the $+$ in the $\pm$ because otherwise the numeerator is negative and you have $e^y>0$ (assuming here that we are just talking about real numbers). So $$ y = \ln(e^y) = \dots $$

$$\text{If }x=\frac{e^y-e^{-y}}2,2xe^y=e^{2y}-1$$

$$\implies e^{2y}-2xe^y-1=0$$ which is a Quadratic Equation in $e^y$

$$\text{ So,}e^y=\frac{2x\pm\sqrt{(2x)^2-4\cdot1\cdot(-1)}}{2\cdot1}=x\pm\sqrt{x^2+1}$$

For real $y,e^y>0\implies e^y=x+\sqrt{x^2+1}$ as $x<\sqrt{x^2+1}$