I'm trying to find the tangent line with respect to an arbitrary point. Let $f(x,y) = x^3 +y^3 -3xy$. We can apply the implicit function theorem here, to solve $y(x)$, in otherwords the equation defines this $y(x)$ implicitly. We can solve for $y'$ since this $y(x)$ is differentiable. By the chain
rule$$\frac {\partial f}{\partial y} = f_y$$$$y'=\frac{-f_x}{f_y}=\frac{-3x^2+3y}{3y^2-3x}$$
provided that $y\neq \sqrt{x}$. Is the equation of the tangent line now simply $$y-y_0 = y'(x)(x-x_0)$$
and can we determine the numbers $y_0,x_0$?
Best Answer
We can just implicitly differentiate $x^3+y^3-3xy=0$ directly; no need to set $f(x,y)$. Doing so gives us
$$ 3x^2+3y^2y'-3y-3xy'=0\implies y'=\frac{y-x^2}{y^2-x} $$
as you found. Now, as you noted, the tangent line to any point $(x_0,y_0)$ that satisfies $x_0^3+y_0^3-3x_0y_0=0$ is
$$y-y_0=\left(\frac{y_0-x_0^2}{y_0^2-x_0}\right)(x-x_0).$$
This is really as simple as it can be, since we shouldn't even try to solve for $y$ explicitly in $x^3+y^3-3xy=0$.