I am given $\Phi=\sum_{i=1}^{k} a_{i}(\mathbf{x}) d x_{i}$ and want to show that $d\Phi=0$ if and only if $$\frac{\partial a_{i}(\mathbf{x})}{\partial x_{j}}=\frac{\partial a_{j}(\mathbf{x})}{\partial x_{i}}$$ This is how I started \begin{aligned} d \Phi &=\sum_{i=1}^{k} d a_{i}(\mathbf{x}) \wedge d x_{i} \\ &=\sum_{i=1}^{k} \sum_{j=1}^{k} \frac{\partial a_{i}(\mathbf{x})}{\partial x_{j}} d x_{j} \wedge d x_{i} \end{aligned}
but I don't know how go from there, can someone please guide me? And how would I use the equation given to me above?
The wedge product $\wedge$ is alternating, meaning that for all $i, j$ with $1 \leq i,j \leq k$ one has $$ dx_j \wedge dx_i = - dx_i \wedge dx_j. $$ Therefore not all the terms in the double sum are linearly independent, some even vanish (when $i = j$): you are allowed to only sum over pairs $(i,j)$ such that $i < j$. $$d\Phi = \sum_{i=1}^{k} \sum_{j=1}^{k} \frac{\partial a_{i}(\mathbf{x})}{\partial x_{j}} d x_{j} \wedge d x_{i} = \sum_{i<j} \left(\frac{\partial a_{i}(\mathbf{x})}{\partial x_{j}} - \frac{\partial a_{j}(\mathbf{x})}{\partial x_{i}} \right) d x_{i} \wedge d x_{j}.$$ On the RHS there are now no redundant terms (indeed the set $\{dx_i\wedge dx_j\}_{i<j}$ forms a basis of the space of $2$-forms), and the components that appear in front of $dx_i \wedge dx_j$ are the actual components of $d\Phi$.
Now you may easily see the equivalence between the two conditions.