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)".

Using the definition,
\begin{eqnarray}
\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}
&=& \lim_{h \to 0} \frac{2(x+h)^2 + 3(h+x) - 2x^2 + 3x}{h}\\
&=& \lim_{h\to 0} \frac{4xh + 3h + h^2}{h}\\
&=& \lim_{h\to 0} \left(4x + 3 + h \right)\\
&=& 4x + 3
\end{eqnarray}
Writing $\lim_{h\to0}\frac{f(x+h) - f(x)}{h} = \lim_{h\to 0} \frac{y - y}{h} = 0$ is not correct.
Since $y$ is a variable that is dependent on $x$, $y$ is function of $x$ and can be written as $y = f(x)$.
