I assume you mean
[tex]\dfrac{\mathrm d}{\mathrm dx}\ln\dfrac{e^x}{e^x-1}[/tex]
The chain rule says this is equivalent to
[tex]\dfrac{\dfrac{\mathrm d}{\mathrm dx}\dfrac{e^x}{e^x-1}}{\dfrac{e^x}{e^x-1}}[/tex]
The numerator can be worked out via the quotient rule:
[tex]\dfrac{\mathrm d}{\mathrm dx}\dfrac{e^x}{e^x-1}=\dfrac{e^x(e^x-1)-e^x\times e^x}{(e^x-1)^2}=-\dfrac{e^x}{(e^x-1)^2}[/tex]
We're left with
[tex]\dfrac{-\dfrac{e^x}{(e^x-1)^2}}{\dfrac{e^x}{e^x-1}}=-\dfrac1{e^x-1}=\dfrac1{1-e^x}[/tex]
