Question: Can someone tell me (or show me), where can I find a proof for the convexity of \[f:\mathbb{R}\_{\geq0}\to\mathbb{R},\quad x\mapsto x\log x\] without using the first or second derivative trick?

We call a map convex if \[f(\lambda x+(1-\lambda)\tilde{x})\leq\lambda f(x)+(1-\lambda)f(\tilde{x})\quad\text{for}~\lambda\in(0,1)\]

Answer: 
Answer to previous version of this question:

This is hair-splitting, but: the first derivative of your function is an increasing function, so your function is convex. 

Hair splitting, because the first derivative trick is not the second derivative trick, and "increasing first derivative" is not exactly "non-negative second derivative", but yet implies convexity, too.
That's what I figured out: Let $\lambda\in (0,1)$ and $x,y\in\mathbb{R}>0$. Then we get
\begin{align\*}
f(\lambda x+(1-\lambda)y)&=[\lambda x+(1-\lambda)y]\log(\lambda x+(1-\lambda)y)\\
\\
&=\lambda x\log(\lambda x+(1-\lambda)y)+(1-\lambda)y\log(\lambda x+(1-\lambda)y).
\end{align\*} 
For the first term of RHS we get 
\[\lambda x\log(\lambda x+(1-\lambda)y)\leq\lambda x\log(x)+\lambda(1-\lambda)y\]
and for the second term we get
\[(1-\lambda)y\log(\lambda x+(1-\lambda)y)\leq (1-\lambda)y\log(y)+\lambda(1-\lambda)x\]
In summary, this results in 
\[f(\lambda x+(1-\lambda)y)\leq \lambda f(x)+(1-\lambda)f(y)+\lambda(1-\lambda)(x+y).\]
The last term is the problem now.