TITLE: Find slopes of tangent lines where $\frac{dy}{dx}$ has removable discontinuity.
QUESTION [6 upvotes]: A colleague of mine wrote a worksheet where she asked students to find the derivative $\frac{dy}{dx}$, where the points $(x,y)$ are defined by the "Heart Curve,"
$$(x^2+y^2-1)^3-x^2y^3=0.$$
She asks students about points where the derivative is well-defined, but I wondered about the points where the derivative isn't well-defined.
A plot of the graph shows sharp points at $(0,\pm1)$, and seems to be a smooth curve at the points $(\pm1,0)$, yet the derivative is undefined at points where $x=0$ or $y=0$.  At points where $x=0$, there should be VERTICAL tangent lines (I have since ran the zoom in on a point test to find this), and at points where $y=0$ there should be one distinct tangent line. It seems slopes here should be $\pm2$.

I have calculated $\frac{dy}{dx}$ as
$$\frac{dy}{dx}=\frac{2xy^3-6x(x^2+y^2-1)^2}{6y(x^2+y^2-1)^2-3x^2y^2},$$
and I feel like there should be a way to remove singularities for $x=0$ or $y=0$ in the expression but I do not know exactly how.
I thought about maybe using partial derivatives and L'Hopital's Rule, but not sure.  My goal is to discuss this with students in a single-variable calculus course, so I'd like a solution that doesn't use partial differentiation.

REPLY [2 votes]: I think it's easier to move the second term to the right to get
$$
(x^2+y^2-1)^3 = x^2y^3
$$
Take the real cube root of both sides to obtain
$$
x^2+y^2-1 = \sqrt[3]{x^2}y
$$
Then differentiate implicitly:
$$
2x+2y\frac{dy}{dx} = \frac{2}{3\sqrt[3]{x}}y + \sqrt[3]{x^2}\frac{dy}{dx}
$$
$$
\left(2y - \sqrt[3]{x^2}\right)\frac{dy}{dx} = \frac{2}{3\sqrt[3]{x}}y - 2x
$$
$$
\frac{dy}{dx} = \frac{\frac{2}{3\sqrt[3]{x}}y - 2x}{2y - \sqrt[3]{x^2}}
$$
and when evaluated at $y = 0$, we obtain
$$
\frac{dy}{dx} = 2\sqrt[3]{x}
$$
which equals $\pm 2$ at $x = \pm 1$, respectively.