Why is the derivative of $y=\frac x y$ different from the derivative of $y^2=x$? I am trying to teach myself calculus but I can't overcome this question...

Since, graphically, equations $y = \frac x y$ and $y^2=x$ are the same, how come $\frac {dy}{dx}$ of the former is $\frac{y}{y^2+x}$ while the derivative of the latter is $\frac{1}{2y}$? Shouldn't they be the same?
 A: Equations do not have "derivatives". Functions have derivatives.
If you "take a derivative" of an equation, what you are really doing is taking the derivative of both sides of the equation, meaning the result must also be an equation.
So when you say that the derivative of $y=\frac xy$ is $\frac{y}{y^2+x}$, that sentence makes no sense.
What you can say is that if you take the derivative of both sides of the equation, you then get
$$\frac{d}{dx} y = \frac{d}{dx}\frac{x}{y}.$$
You can then simplify this expression, using the rule of differentiating fractions, to get
$$y' = \frac{y - y'\cdot x}{y^2}$$
and from here, you can get that
$$y' = \frac{y}{y^2+x}.$$

On the other hand, you can similarly manipulate the expression $y^2=x$ to get that $$y' = \frac{1}{2y}$$
and these two expressions are in no way contradictiory. In fact, both relations are true, and you can actually morph one into the other since
$$\begin{align}\frac{1}{2y} &=\frac{1}{y+y}
\\&=\frac{1}{y + \frac{x}{y}}
\\&=\frac{1}{\frac{y^2+x}{y}}
\\&=\frac{y}{y^2+x}\end{align}$$
where I used the fact that $y=\frac xy$ to get from the first row to the second.
A: That's a perfectly reasonable question. I like to write implicit derivatives like this one in terms of the differential and variations in the independent variable, with $\delta y = dy\,\delta x$ denoting the first-order change in $y(x)$ due to the variation $\delta x$ in $x$.
If we differentiate both sides of both equations we get
$$dy\,\delta x = \frac{1}{y(x)} \delta x - \frac{x}{y(x)^2} \, dy\,\delta x$$
and
$$2y(x) dy\,\delta x = \delta x.$$
Note that this procedure is only well-defined when the conditions of the implicit function theorem hold, i.e., away from $x=0$.
Solving for $\delta y$ in both cases,
$$\left(1 + \frac{x}{y(x)^2}\right)\delta y = \frac{1}{y(x)} \delta x$$
$$\delta y = \frac{1}{2 y(x)}\delta x.$$
Plugging in $y = \sqrt{x}$ (or $y = -\sqrt{x}$, depending on which branch you're on) you will see that both expressions are equal.
