Conditions For a Vector Function to be a Complete Differential

Theorem
If
$\mathbf{F} = (F_1 , F_2 , F_3 )$
is a vector field, a necessary and sufficient condition for
$F_1 dx + F_2 dy + F_3 dz$
to be a complete differential is that
$curl \mathbf{F} = 0$

Proof
Assume
$F_1 dx + F_2 dy + F_3 = \frac{\partial \phi}{\partial x} dx + \frac{\partial \phi}{\partial y} dy + \frac{\partial \phi}{\partial z} dz = d \phi$
is an exact differential then
$F_1 = \frac{\partial \phi}{\partial x}, \: F_2 = \frac{\partial \phi}{\partial y}, \: F_3 = \frac{\partial \phi}{\partial z}$

Therefore
$\mathbf{F} = \frac{\partial \phi}{\partial x} \mathbf{i} + \frac{\partial \phi}{\partial y} \mathbf{j} + \frac{\partial \phi}{\partial z} \mathbf{k} = d \phi$

Now use the identity
$curl grad \phi =0$
with
$grad \phi = \mathbf{F}$
to obtain
$curl \mathbf{F} =0$

Conversely, if
$curl \mathbf{F} =0$
there exists
$\phi$
such that
$\mathbf{F} = grad \phi$

Then
\begin{aligned}F_1 dx+ F_2 dy + F_3 dz &= \mathbf{F} \cdot d \mathbf{r} \\ &= (F_1 \mathbf{i} + F_2 \mathbf{j} + F_3 \mathbf{k}) \cdot (dx \mathbf{i} + \mathbf{j} + \mathbf{k} ) \\ &=(\frac{\partial \phi}{\partial x} \mathbf{i} + \frac{\partial \phi}{\partial y} \mathbf{j} + \frac{\partial \phi}{\partial z} \mathbf{k}) \cdot (dx \mathbf{i} + \mathbf{j} + \mathbf{k} ) \\ &= \frac{\partial \phi}{\partial x} dx + \frac{\partial \phi}{\partial y} dy + \frac{\partial \phi}{\partial z} dz \\ &= d \phi \end{aligned}