I was solving the following ODE
$$ y \cos(xy) + x \cos(xy) y' = 0$$
My initial intuition to solving it is to observe that it can be factored as:
$$ \cos(xy) (y + xy') = 0$$
So over $x,y$ that don't satisfy $\cos(xy) = 0$ we have
$$ y + xy' = 0$$
And this is an exact eqation which has solutions of the form
$$ xy = C$$
Now the video series that the ODE originated from takes a different approach. It notices that $$\frac{\partial}{\partial y} [y \cos(xy)] = \cos(xy)- xy \sin(xy) , \frac{\partial}{\partial x} [x\cos(xy)] = \cos(xy) - xy \sin(xy) $$
So therefore the solutions are of the form $ \sin(xy) = C$
clearly my approach doesn't agree with the classical approach, but my only difference was dividing out a term which is 0 in a set non-dense in $\mathbb{R}^2$ so I don't understand why my answer differs so much. And my intuition that $xy$ should be constant for all points where $\cos(xy) \ne 0$ can't possibly be true so I feel i'm deeply missing some intuition here.
Indeed, $xy=C$ and $\sin(xy)=C$ characterize the same family of curves. For example $\sin(xy)=0$ is the curves $xy=k\pi$, $k\in\mathbb{Z}$.