$$x^3+y^3=6xy\tag{1}\label{eq1}$$
$$\frac{dy}{dx}=\frac{\left(2y-x^{2}\right)}{y^{2}-2x}\tag{2}\label{eq2}$$
I embed the derivative $(2)$ into a 3D space by including variable $z$:
$$z=\frac{\left(2y-x^{2}\right)}{y^{2}-2x}\tag{3}\label{eq3}$$
I suppose this embedded function $(3)$ gives me the set of "all possible derivatives" of function $(1)$.
To find the set of all "actual derivatives" I have to:
- Embed function $(1)$ into 3D space
$$z=x^{3}+y^{3}-6xy\tag{4}\label{eq4}$$ - Intersect function $(4)$ with function $(3)$.
$$\frac{\left(2y-x^{2}\right)}{y^{2}-2x}=x^{3}+y^{3}-6xy\tag{5}\label{eq5}$$
Then I project the intersection in (5) back to 2D space.
Projection of function $(5)$ onto 2D space
Is this projection the set of all $(x,y)$ pairs of function $(1)$ that have a derivative? What can be inferred from projection about function $(1)$?
Best Answer
The derivative of implicit function $R(x,y) = 0$ is $$\frac{d y}{d x} = - \frac{\partial R / \partial x}{\partial R / \partial y}$$ and in this case, $R(x,y) = x^3 + y^3 - 6 x y$, and $$\frac{\partial R}{\partial x} = 3 x^2 - 6 y, \quad \frac{\partial R}{\partial y} = 3 y^2 - 6 x$$ therefore $$\frac{d y}{d x} = \frac{ 3 x^2 - 6 y }{ 3 y^2 - 6 x } = \frac{2 y - x^2}{y^2 - 2 x} = \frac{ x^2 - 2 y }{ 2 x - y^2 }$$ as you have already calculated.
The equation $x^3 + y^3 - 6 x y = 0$ is an implicit curve, and the derivative above tells us its slope ($d y / d x$) at a point $(x, y)$.
Because of this, your intersection does not tell anything about that implicit curve or its derivative, and just describes the points where $(x^2 - 2 y) / (2 x - y^2) = x^3 + y^3 - 6 x y$, i.e. it is just the implicit curve $$(x^2 - 2 y) - (2 x - y^2)(x^3 + y^3 - 6 x y) = 0$$
Another approach to examining the implicit curve would be to rewrite it in parametric form. We do this by substituting $y = t x$ to the implicit form, $$x^3 + y^3 - 6 x y = 0 = x^3 + t^3 x^3 - 6 t x^2$$ and then solving for $x$. First it simplifies to $$x^2 ( x + x t^3 - 6 t ) = 0$$ and ignoring the double root $x^2 = 0$, we get $$x + x t^3 - 6 t = x (1 + t^3) - 6 t = 0 \quad \iff \quad x = \frac{6 t}{1 + t^3}$$ For $y$, we do the same substitution for $x$, i.e. $x = y / t$, and then solve for $y$: $$\frac{y^3}{t^3} + y^3 - \frac{6 y^2}{t} = 0$$ which simplifies to $$y^2 \left(\frac{y}{t^3} + y - \frac{6}{t}\right) = 0$$ and again ignoring the double root $y^2 = 0$, we get $$\frac{y}{t^3} + y - \frac{6}{t} = 0 = y \frac{t^3 + 1}{t^3} - \frac{6 t^2}{t^3} \quad \iff \quad y = \frac{6 t^2}{t^3 + 1}$$ In other words, we can parametrize the implicit curve using $t$ as $$\begin{cases} \displaystyle x(t) = \frac{6 t}{1 + t^3} \\ \displaystyle y(t) = \frac{6 t^2}{1 + t^3} \\ \end{cases} $$ (This is the folium of Descartes with $a = 2$, by the way.)
The tangent (vector) of the curve at point $t$ is $$\begin{cases} \displaystyle \frac{d x(t)}{d t} = \dot{x}(t) = \frac{6 - 12 t^3}{(1 + t^3)^2} \\ \displaystyle \frac{d y(t)}{d t} = \dot{y}(t) = \frac{12 t - 6 t^4}{(1 + t^3)^2} \\ \end{cases}$$ and the slope $d x / d y$ at that point $t$ is $$z(t) = \frac{\dot{x}(t)}{\dot{y}(t)} = \frac{2 t^3 - 1}{t^4 - 2 t}$$ Now, if you want to visualize the slope of the implicit function in 3D, you can now just draw the 3D parametric curve $\vec{p}(t) = \big( x(t), y(t), z(t) \big)$ (noting that it does not exist at points where the curve is parallel to the $y$ axis, as then the slope is infinite and $z(t)$ involves division by zero; this occurs at $t=0$ and $t=2^{1/3} \approx 1.259921$.)
Interactively, or to illustrate tangents at specific points, you can first draw the $\big(x(t), y(t)\big)$ curve, then draw one or more tangent vectors at $t$ with lines or arrows from $\big(x(t), y(t)\big)$ to $\big(x(t)+\dot{x}(t), y(t)+\dot{y}(t)\big)$.