calculus - Function as a "constant of integration"

I'm reading a book Differential Equations with Applications and Historical Notes, 3rd edition, specifically section 8 about exact equations. The author is trying to prove that iff $\partial M/\partial y = \partial N/\partial x$ then equation
\begin{equation}
M(x,y)dx + N(x,y)dy = 0
\end{equation}

is exact differential equation.
At some point we integrate equation
\begin{equation}
\frac{\partial f(x,y)}{\partial x} = M(x,y)
\end{equation}
to get
\begin{equation}
f(x, y) = \int M(x,y)dx + g(y)
\end{equation}
The author states that function $g(y)$ appears as a constant of integration because if we take derivative of both sides with respect to $x$, $g(y)$ would disappear because it doesn't depend on $x$.
That's the part that I have trouble with, $y$ is a dependent variable and $x$ is independent variable so wouldn't derivative of $g(y)$ with respect to $x$ be
\begin{equation}

\frac{d\,g(y)}{dy} \frac{dy}{dx}
\end{equation}
and not $0$ ?

Answer

This is a common poorly written part in differential equations textbooks, because they don't want to spend time discussing differential forms.

At this point we forget that $y$ depends on $x$. Of course then the equation $M(x,y)dx+N(x,y)dy=0$ looks weird, and indeed it's wrong. What is meant there is that if we have a dependence of $x$ and $y$, a curve on $x$-$y$ plane, denoted $\gamma$, then the pullback of $M(x,y)dx+N(x,y)dy$ on $\gamma$ is $0$. For example, if we can parametrize $\gamma$ by $x$ (i.e. we can write $y$ as a function of $x$), then this condition says $\frac{dy}{dx} = -\frac{M(x,y)}{N(x,y)}$. That's why we want to find such $\gamma$.

The exactness condition means that $df=M(x,y)dx+N(x,y)dy$. Then the level sets of $f$, $\{(x,y)|f(x,y)=c\}$, give us such $\gamma$'s. Note that exactness follows from closeness on simply connected domains.

So, one can separate this problem into two stages, where $x$ and $y$ are independent, and then were we look for a required dependence.

Alternatively, instead of using differential forms, one can think of $(N,M)$ as a vector field on $x$-$y$ plane perpendicular to $\gamma$'s, the level sets of $f$, gradient of which is $(N,M)$.