The function $\frac{x}{e^x-1}$ has limit $1$ at $0$. It has derivative defined everywhere except at the origin. At the origin, the derivative exists and it is $0$. The limit of the second derivative is positive, but it is not a minimum point. It is not a turning point.
Does that mean the function $\frac{x}{e^x-1}$ is not differentiable at $0$?
I assume you're considering $$ f(x)=\begin{cases} \dfrac{x}{e^x-1} & \text{if $x\ne0$}\\[6px] 1 & \text{if $x=0$} \end{cases} $$ It is an everywhere continuous function.
The function is obviously differentiable for $x\ne0$, with $$ f'(x)=\frac{e^x-1-xe^x}{(e^x-1)^2} $$ Since $$ \lim_{x\to0}f'(x)= \lim_{x\to0}\frac{(1+x+x^2/2+o(x^2))-1-x(1+x+o(x))}{(x+o(x))^2}= -\frac{1}{2} $$ the function is also differentiable at $0$, by l’Hôpital’s theorem.
