$x\mapsto \sum_{n=1}^{\infty}ne^{-nx}$ is continuous but not uniform continuous on $(0,+\infty)$. 
Prove that $f(x)=\sum_{n=1}^{\infty}ne^{-nx}$ is continuous but not uniform continuous on $(0,+\infty)$.

Is there any good ways to deal with this kind of functions?

Yes. To prove the continuity we work locally and we use the uniform convergence:
$$ne^{-nx}\le ne^{-n a},\quad\forall x\ge a$$
and the series $\sum_n ne^{-na}$ is convergent because $ne^{-na}=_\infty o\left(\frac1{n^2}\right)$  hence we have the uniform convergence of the given series on every interval $[a,+\infty),\;a>0$ and since the functions $x\mapsto ne^{-nx}$ are continuous we conclude the desired result.
This series isn't uniformly convergent on $(0,\infty)$ because $\lim\limits_{x\to0}f(x)$ doesn't exist.
Edit We can prove that the series isn't uniformly convergent using:
$$\sup_{x>0}\sum_{k=n+1}^\infty ne^{-nx}\ge \sup_{x>0}\sum_{k=n+1}^{2n} ne^{-nx}=2n^2\xrightarrow{n\to\infty}\infty$$
