Answer
$$ f'(x)=
\frac{2 e^{x}-e^{x+1}-2 x e^{x}-1}{\left(2 e^{x}-1\right)^{2}}.$$
Work Step by Step
Recall that $(e^x)'=e^x$
Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$
Since we have
$$ f(x)= \frac{e^{x+1}+x}{2e^x-1}$$
then the derivative $ f'(x)$, using the chain and quotient rules, is given by
\begin{align*}f'(x)&=\frac{(2e^x-1)(e^{x+1}+x)'-(e^{x+1}+x)(2e^x-1)'}{(2e^x-1)^2} \\
&=\frac{(2e^x-1)(e^{x+1}+1)-(e^{x+1}+x)(2e^x)}{(2e^x-1)^2} \\
&=
\frac{2 e^{x}-e^{x+1}-2 x e^{x}-1}{\left(2 e^{x}-1\right)^{2}}.
\end{align*}