$P(x,y)dx+Q(x,y)dy$ is an exact differential iff $ \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$, while $ A_1 (x,y,z) dx + A_2 (x,y,z) dy + A_3 (x,y,z) dz$ is an exact differential iff $\nabla \times \mathbf{A} = \mathbf{0}$ where $\mathbf{A} = A_1 \mathbf{i} + A_2 \mathbf{j} + A_3 \mathbf{k}$. I wonder if exists a simple general rule that summarizes necessary and sufficient conditions for \begin{equation} f_1 (x_1,\dots,x_n) dx_1 + \dots + f_n (x_1,\dots,x_n) dx_n \end{equation} to be exact. I fear a complete answer to this needs a book, but leaving away a proofs and supposing that functions are very well behaved?
Edit
I guess (it's correct?) that these four claims are equivalent:
$\quad$ i) $\frac{\partial f_i}{\partial x_j} = \frac{\partial f_j}{\partial x_i} \quad i \neq j \quad$ everywhere
$\quad$ ii) $df$ is exact ($\exists f \, : \, df= \sum_{x_i} \frac{\partial f}{\partial x_i} dx_i$)
$\quad$ iii) Integral between two point $(x_1,\dots,x_n)$ is indipendent from path ($\int df$ is path independent)
$\quad$ iv) Integral on a closed path is zero ($\oint_P df = 0 \,\, \forall P $)
(iii) $\Leftrightarrow$ (iv) is simple, (ii) $\Rightarrow$ (iii) (and so (ii) $\Rightarrow$ (iv)) too is simple. It is not hard to show that ((i)$\wedge$(iii))$\Rightarrow$(ii). If it were possible to show that (iii)$\Rightarrow$(i) we would have the fully equivalence of (ii), (iii) and (iv) and we only would need one of (i)$\Rightarrow$(ii), (i)$\Rightarrow$(iii) or (i)$\Rightarrow$(iv) to end the job.
Is it possible to follow a such course? How to prove (iii)$\Rightarrow$(i) and one between (i)$\Rightarrow$(ii), (i)$\Rightarrow$(iii) or (i)$\Rightarrow$(iv)? But first of all: are the four claims equivalent?
You are ignoring issues with singularities, but, yes, if the curl is everywhere zero in $\Bbb R^3$, then $\mathbf A$ is conservative, i.e., the $1$-form is exact.
In $n$ dimensions, the analogous statement is that if $$\frac{\partial f_i}{\partial x_j} = \frac{\partial f_j}{\partial x_i} \quad\text{for all }i\ne j,$$ everywhere in $\Bbb R^n$ (or in a simply-connected region), then the $1$-form $\sum f_i\,dx_i$ is exact.