What's the rule for knowns and unknowns when dealing with derivatives So a rule of thumb when doing basic algebra is you must have as many equations as you have unknowns. For example:
$0=4x+6y^2$
$3x=2\sqrt{y}$
You have two equations and two unknowns and thus can solve for both x and y
However I don't know how this rule (if it even applies) works if you have a differential equation instead for example
$0=4x+6y^2$
$\frac{dy}{dx}=2x$
Does the second equation still count as one equation or does it introduce additional information that's needed to solve the problem? Does the general (rule 1 equation for 1 unknown) still work for all cases? What about higher order differential equations?
 A: In this case, you can still use substitution:
$$2x = \frac{dy}{dx} \implies 0=2\frac{dy}{dx}+6y^2.$$
Now solve.
More generally, this is a more complicated type of system known as a differential-algebraic equation (DAE).
We may view it in vector form as
$$\mathbf{F}\left(\frac{dy(x)}{dx}, y(x), x\right) \stackrel{\textrm{def}}{=} \begin{pmatrix} 4x+6y^2 \\ \frac{dy(x)}{dx}-2x \end{pmatrix} = \mathbf{0}.$$
What you have conceptualized is known as a semi-explicit DAE of index 1. These types of DAEs include a purely algebraic term, and a differential equation term.
Solvability is more restrictive than for systems of algebraic equations. For index 1 DAEs, we must have that the purely algebraic term is solvable in terms of $y$. At this point, I'm just repeating things that are already present on the Wikipedia entry, but the system you have found belongs to a family of problems that can actually be quite challenging to solve.
A: Once integrated, the differential equation becomes an ordinary equation but that includes an integration constant, i.e. a new unknown.
In your example,
$0=4x+6y^2$
$\frac{dy}{dx}=2x\implies y=x^2+C$
