Hi, nfxgosu!
You just have to solve for 'x':
[tex]f(x)=y=\frac{e^x}{1+2e^x}[/tex]
[tex](1+2e^x)y=e^x[/tex]
[tex]y+2ye^x-e^x=0[/tex]
[tex]e^x(2y-1)=-y[/tex]
[tex]e^x=-\frac{y}{2y-1} /ln[/tex]
[tex]x=\ln\frac{y}{1-2y}[/tex]
So, for the inverse function we have:
[tex]\overline{f}(x)=\ln\frac{x}{1-2x}[/tex]
Best wishes, Marine