What does it mean when we say $\frac{dy}{dx} = f(x)$? If we have an equation that's true for all $(x,y)$ in a set $S$, for example:
$y = 2x^2 + 3x$
We say that its derivative is:
$\frac{dy}{dx} = 4x + 3$
However, if we use the definition of the derivative on the left half $f(x) = y$, we get:
$\lim_{h\to0}\frac{f(x+h) - f(x)}{h} = \lim_{h\to 0} \frac{y - y}{h} = 0$
This would mean that either: 


*

*Whenever $y = 2x^2 + 3x \rightarrow 0=4x+3$, which is false

*$\frac{dy}{dx}$ doesn't always mean "the derivative of $f(x) = y$"


The same problem also happens with implicit differentiation: 
$x^2 + y^2 = 1$
$2x + \frac{dy^2}{dx} = 0$
$2x + lim_{h\to0}\frac{y^2 - y^2}{h} = 0$
$2x = 0$  (not equivalent to $x^2+y^2=1$)
All I did to get that result was to use the limit definition of the derivative on $\frac{dy}{dx}$ instead of leaving it as it is and isolating it.
What am I doing wrong?
Also, by $x$ and $y$, I am referring to variables in a two-variable equation. I'm not using "y" to mean "f(x)".

The expression $\frac{dy}{dx}$ is read "the derivative of the variable $y$ with respect to the variable $x$". Actually, this refers to the derivative of the function that relates the variables.
If $y=2x^2+3x$ and $y=f(x)$, then $f(x)=2x^2+3x$ and $f(x+h)=2(x+h)^2+3(x+h) \neq y$. This is your mistake.
The definition is for the derivative of a function, not of a variable.
$y=f(x)$ and $y=f(x+h)$ are different ways of relating the variables, and that is what causes a zero result (if both hold for all small values of $h$, this would mean that $y$ is constant with respect to $x$, so the derivative is zero).
You have the right to say that $x$ and $y$ are variables and that $y$ is not only an expression for $f(x)$, but it must be clear what function you are trying to differentiate. And remember that $y=f(x)$ is an equation relating the variables as much as $x^2+y^2=1$ is
