There is a specific theorem that I would like to make sure I understand correctly.
Here is the statement of the theorem as it appears in Apostol's Calculus, Volume II, section 10.19 (on application of line integrals to exact differential equations of first order):
Theorem 10.7. Assume that the differential equation
$$P(x,y)dx+Q(x,y)dy=0\tag{10.14}$$
is exact in an open connected set $S$, and let $\varphi$ be a scalar field satisfying
$$\frac{\partial\varphi}{\partial x}=P$$
$$\frac{\partial\varphi}{\partial y}=Q$$
everywhere in $S$. Then every solution $y=Y(x)$ of (10.14) whose graph lies in $S$ satisfies the equation $\varphi[x,Y(x)]=C$ for some constant $C$. Conversely, if the equation
$$\varphi(x,y)=C$$
defines $y$ implicitly as a differentiable function of $x$, then this function is a solution of the differential equation (10.14).
Note that a differential equation like that in (10.14) is said to be exact in $S$ if when we associate a vector field $V(x,y)=P(x,y)\hat{i}+Q(x,y)\hat{j}$ with it, the vector field is the gradient of some scalar field $\varphi$.
Here is my interpretation of what this theorem says.
If we have a differential equation that
we can write in the form 10.14, ie $P(x,y)dx+Q(x,y)dy=0$ and
the vector $\langle P(x,y), Q(x,y)\rangle$ is the gradient of a scalar field $\varphi$
then any solution $y(x)$ of the differential equation must satisfy $\varphi(x, y(x))=0$.
My question is if the following paragraph is correct
In other words, if the righthand side of 10.14 is the derivative of a scalar field $\varphi$ along a path $(x,y(x))$ then the fact that we are equating this derivative to zero means that $\varphi$ is constant along this path, ie $\varphi(x,y(x))=C$ for some $C$.
Here is a generic picture I drew to depict this latter paragraph
In this picture, the derivative of $\varphi(x,y(x)$ relative to $x$ equals zero and is our original differential equation.