I had problems with finding the equation of a tangent line (using implicit differentiation) to the curve $x^3 -xy + y^3= 0$ at the point $(0,0)$.
I used implicit differentiation to find the derivative (slope of tangent line) of $y$ with respect to $x$, but in substituting $(0,0)$ there yields $0/0$, which is undefined.
Please answer the question with mathematical detail.
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$\begingroup$Via implicit differentiation, treating $y$ as a function of $x$:$$ 3 x^2 - x y' - y + 3 y^2 y' = 0 \text{.} $$So$$ (-x +3y^2) y' = y - 3x^2 $$and then $$ y' = \frac{y - 3 x^2}{3y^2 - x} = 9 y^2 + 3x + \frac{y(27 y^3 - 1)}{3y^2 - x} \text{.} $$
(Just plugging in $y = x = 0$ gives us the indeterminate form, $\left[ \frac{0}{0} \right]$, so we expect to use limits to resolve what is happening.)
It is easy enough to inspect the discriminant of the cubic (in $y$ or $x$, since they appear symmetrically in the equation), $-27x^6 + 4x^3$, to discover there is only one solution for any $x < 0$ and likewise for any $y < 0$. Let us inspect the solution for small negative $x$, $x = - \varepsilon$. Then the given equation is $$ y^3 + \varepsilon y + \varepsilon^3 = 0 \text{.} $$In the limit as $\varepsilon \rightarrow 0$, we only have the root $y = 0$. Taking $y = 0$ in the previous expression for the derivative, we get $0 + 3x + 0$ and letting this $x \rightarrow 0^-$, we find that the derivative for the solution approaching $(0,0)$ from the left approaches zero from below. The equation of this tangent line is $y = 0$.
Repeating the analysis for small negative $y$, we find the $x = 0$. Putting that in the expression for the derivative obtains $9 y^2 + 0 + \frac{27 y^3 - 1}{3y}$. As $y \rightarrow 0^-$, we find the derivative is undefined. That is, the tangent line is vertical, $x = 0$.
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