Let $[a_{i,j}(x_1,\ldots,x_n)]$ be a skew-symmetric $n\times n$ matrix of functions $a_{i,j}\in C^\infty(\mathbb{R}^n)$. Show that the vector field $$v=\sum\left(\dfrac{\partial}{\partial x_i}a_{i,j}\right)\dfrac{\partial}{\partial x_j}$$ is divergence-free.
It looks like there's a typo somewhere in this problem. For one thing, it looks like there should be a double summation, one over $i$ and the other one over $j$. Should it be $$\sum_{j=1}^n\left(\sum_{i=1}^n\dfrac{\partial}{\partial x_i}a_{i,j}\right)\dfrac{\partial}{\partial x_j}?$$
The definition of a vector field $v$ to be divergence-free is that the divergence $\sum_{j=1}^n\dfrac{\partial v_j}{\partial x_j}=0$.
For your first point, the answer is yes, it is a double sum. You'll see a lot of shorthand summing notation while you study differential geometry! Read this as $$ \sum_{ij} \left(\frac{\partial}{\partial x_i} a_{ij}\right) \frac{\partial}{\partial x_j} $$
Hint for the proof of divergence free: Mixed partial derivatives commute on functions in $C^{\infty}(\mathbb{R})$ and, by assumption, $a_{ij} = -a_{ji}$.