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