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