In the following procedure, how is the function (implicitly) chosen? Given $x^3+y^3-3xy=0$ we differentiate with respect to $x$ on both sides, yielding $3x^2+3y^2\frac{dy}{dx} -3x\frac{dy}{dx} - 3y = 0$, and rearrange to get
$$
\frac{dy}{dx} = \frac{y-x^2}{y^2-x},\quad y^2\neq x
$$
The original relation $x^3+y^3-3xy=0$ implicitly defines any number of functions $y(x)$, since for $0 \leq x\leq 2^{2/3}$ it doesn't uniquely determine a value. Nevertheless this process somehow "selects" one of these (It seems to be the lower one). How does this happen? Does it always happen with implicit functions such as these? Why was this specific function selected, and by what process? Another valid function could be one that in the interval $0 \leq x\leq 2^{2/3}$ that varied between the upper and lower ones at rational and irrational points, and any other number of configurations.
 A: When you differentiate, you're assuming the function is differentiable, so this method can only produce differentiable solutions (and in particular, only continuous solutions, since differentiability implies continuity). The condition $y^2 \neq x$ means that $x \neq 0$ and $x \neq 2^{2/3}$, so the solutions can be considered separately on the intervals $(-\infty, 0)$, $(0, 2^{2/3})$, and $(2^{2/3}, \infty)$. On $(-\infty, 0)$ and $(2^{2/3}, \infty)$, the solution is unique, so there's no choice to be made.
For $0 < x < 2^{2/3}$, there are three choices of implicit function that are continuous. (That there are at most three follows from the fact that the curve is defined by a cubic equation. That there are at least three can be observed by graphing the curve.) All three of these functions satisfy the differential equation
$$\frac{dy}{dx} = \frac{y - x^2}{y^2 - x}.$$
When we consider only the interval $(0, 2^{2/3})$, there is nothing privileging any one of these three choices.
What makes one choice special in this case is how it behaves at the boundary points. All three branches intersect at $(0, 0)$, so the function extends to a continuous function on $(-\infty, 2^{2/3})$ regardless of which of the three branches we choose. However, only the middle branch gives differentiability at $x = 0$. At the right endpoint, only the lower branch allows us to extend to a continuous function on $(0, \infty)$.
So, if you just care about continuity, the lower branch is the "nicest" choice of the three, because it gives us a continuous function on the whole real line. But if you also want the implicit function to extend to something differentiable at the boundary points, there's no perfect choice: the middle branch gives differentiability at $x = 0$ but not $x = 2^{2/3}$, while the lower branch gives differentiability at $x = 2^{2/3}$ but not $x = 0$. And if you don't care about the boundary points, all three are equally reasonable.
