Question on exact differential equations: how do we know $\frac{dy}{dx} = 0$?

Question

My text states that an exact differential equation is one for which $$ du \triangleq M(x,y)dx + N(x,y)dy = 0. $$ And from this it follows that $$ \frac{\partial u}{\partial x} = M(x,y)~\text{ and }~\frac{\partial u}{\partial y} = N(x,y). $$ which seems to imply that $$ \frac{dy}{dx} = \frac{dx}{dy} = 0. $$ I assume this last point follows from the definition of $du$ above but I don't see how.

Answer 1

Never mind. I sloppily read $\frac{\partial u}{\partial x}=M(x,y)$ as $\frac{du}{dx}=M(x,y)$ which caused my confusion. Because in this case, we would have \begin{align} du = M(x,y)dx + N(x,y)dy ~~\text{ and }~~~ \frac{du}{dx} = M(x,y) \end{align} and therefore that $$ N(x,y)dy = 0 ~\text{ which implies that }~ N(x,y)\frac{dy}{dx} = 0. $$ And assuming $N(x,y) \not= 0$, this means $\frac{dy}{dx} = 0$.