I'm having trouble/mistake proving this since when I try to calculate the hessian, the partial derivatives of i,j elements are negative.
This is my current direction and I don't know how to continue:
Since $f(\boldsymbol{x})$ is a continuously differentiable function, the Hessian matrix is symmetric. Now, we can compute the diagonal and off-diagonal entries of the Hessian matrix as follows:
$$ \frac{\partial^{2} f}{\partial x_{i}^{2}} = \frac{n-1}{n^{2} x_{i}^{2n-1}} \prod_{j\neq i}^{n} x_{j}^\frac{1}{n} $$
and
$$ \frac{\partial^{2} f}{\partial x_{i} \partial x_{j}} = - \frac{1}{n^{2}x_{i}^{\frac{n-1}{n}} x_{j}^{\frac{n-1}{n}}} \prod_{k \neq i,j}^{n} x_{k}^{\frac{1}{n}} $$
Any help would be appreciated
From your last result, we have $$ \frac{\partial^{2} f}{\partial x_{i}^{2}} = \frac{n-1}{n^{2} x_{i}^{\color{red}2}} \prod_{j=1}^{n} x_{j}^\frac{1}{n} $$ $$ \frac{\partial^{2} f}{\partial x_{i} \partial x_{j}} = - \frac{1}{n^{2}x_{i} x_{j}} \cdot \prod_{k =1}^{n} x_{k}^{\frac{1}{n}} $$
The term $\frac{1}{n^2}\prod_{k =1}^{n} x_{k}^{\frac{1}{n}}$ is positive, so, it suffices to prove: the matrix $\mathbf{M}=(m_{ij})_{1\le i,j\le n}$ defined by $$m_{ij} = \begin{cases} \frac{n-1}{x_i^2} & \text{for }i=j\\ \frac{1}{x_i x_j} & \text{for }i\ne j\\ \end{cases}$$ is positive definite for the convexity.
For any vector $\mathbf{A} = (a_i)_{1\le i\le n}$, if we can prove that $\mathbf{A}^T \mathbf{M} \mathbf{A} \ge0$ then we can prove $\mathbf{M}$ is a positive definite matrix. Indeed, we have $$ \begin{align} \mathbf{A}^T \mathbf{M} \mathbf{A} &= \sum_{i=1}^n \left(\frac{n-1}{x_i^2}\cdot a_i^2\right) -\sum_{1\le i , j \le n}\left( \frac{1}{x_ix_j}\cdot a_ia_j\right) \\ &= \sum_{i=1}^n (n-1)\left(\frac{a_i}{x_i}\right)^2 -2\sum_{1\le i < j \le n}\left( \frac{x_i}{a_i}\cdot \frac{x_j}{a_j}\right) \\ &=\sum_{1\le i < j \le n} \left(\left(\frac{a_i}{x_i}\right)^2-2\frac{x_i}{a_i}\cdot \frac{x_j}{a_j} + \left(\frac{a_j}{x_j}\right)^2\right)\\ &=\sum_{1\le i < j \le n} \left(\frac{a_i}{x_i} - \frac{a_j}{x_j}\right)^2 \ge 0 \end{align}$$
By consequence, $\mathbf{M}$ is a positive definite matrix and so is the Hessian matrix.
We can conclude that the function is convex.
Q.E.D