for now I've got the following: set $y\_i=e^{x\_i}\quad\Rightarrow \frac{\partial y\_i}{\partial x\_i}=y\_i\quad \Rightarrow$

$\nabla\_x f=\frac{1}{\sum\_{i=1}^{n}{y\_i}}[y\_1\; \dots\; y\_n]^\top\quad\Rightarrow$

$\nabla\_x^2f\_{ij}=\begin{cases} \frac{1}{\big(\sum\_{k=1}^{n}{y\_k}\big)^2}y\_i(\sum\_{k=1}^{n}{y\_k}-y\_i), &\forall\; i=j\\ -\frac{1}{\big(\sum\_{k=1}^{n}{y\_k}\big)^2}y\_iy\_j, & \forall\; i\neq j \end{cases}$

Now I need to show that $\nabla\_x^2f$ is positive semidefinite, but I'm stuck here.
Also tried first order conditions and direct proof by definition of convexity, but also stuck there.