Question: $f$ is a function from $\Bbb R$ to $(0,2)$ defined as 
$$f(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}+1.$$
The function $f$ is invertible and I want to find its inverse. 

I tried using methods like taking $\ln$ on both sides but they don't seem to work.

Answer: 
Here is my go at it. First, observe that we have the two hyperbolic trig functions
$$ \mathrm{sinh}(x)=\dfrac{e^{x}-e^{-x}}{2}, \qquad \mathrm{cosh}(x)=\dfrac{e^{x}+e^{-x}}{2}$$
which defines $\mathrm{tanh}(x):=\mathrm{sinh}(x)/\mathrm{cosh}(x)$. It follows form this that your function $f$ may be written as $f(x)=\mathrm{tanh}(x)+1$. From this, one knows that the inverse function of the hyperbolic tangent is well defined and equal to 
$$\mathrm{artanh}(x)=\frac{1}{2}\ln \left(\frac{1+x}{1-x}\right)$$
with domain $(-1,1)$. Thus,
$$\mathrm{tanh}(x)=f(x)-1\qquad\Rightarrow\qquad f^{-1}(x)=\frac{1}{2}\ln \left(\frac{x}{2-x}\right).$$
If your are not satisfied with my use of hyperbolic trig functions, you can always "reverse engineer" the inverse.

I hope this helps!
$$y=f(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}+1$$
$$y-1=\frac{e^x-e^{-x}}{e^x+e^{-x}}$$
$$y-1=\frac{e^{2x}-1}{e^{2x}+1}$$
$$ye^{2x} +y -e^{2x}-1=e^{2x}-1$$
$$e^{2x}(y-2)=-y$$
$$e^{2x}=\frac{y}{2-y}$$
Taking log 
$$2x=\log\left(\frac{y}{2-y}\right)$$
$$x=\frac12\log\left(\frac{y}{2-y}\right)$$
Now$$ f^{-1}(x)=\frac12\log\left(\frac{x}{2-x}\right)$$