Is the sum of this series a differentiable function?

You can do some simplifications first. The factor $\frac 1 x$ has no effect on differentiability, so drop that. Next note that $1-\frac 1 {e^{x/n}}=1-e^{-x/n}$. Alos recall that $1-e^{-t} \leq t$ for all $t \geq 0$. It is now obvious that the series is uniformly convergent (by comparison with $\sum \frac 1 {n^{2}})$. If you drop $\frac 1 x$ and differentiate the series you will get $\sum\frac x {n^{2}} e^{-x/n}$. Use the fact that $e^{-x/n} \leq 1$ to conclude that the differentiated series also converges uniformly for bounded $x$. Hence the sum of the original series is a differentiable function (as I explained to you in your previous post).