<s>[INST] Find slopes of tangent lines where $\frac{dy}{dx}$ has removable discontinuity. 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.
 [/INST] This solution works with implicit differentiation. The equation of the heart-curve is
$${({x^2} + {y^2} - 1)^3} - {x^2}{y^3} = 0\tag{1}$$
Using implicit differentiation, one can obtain the derivative to be
$${{dy} \over {dx}} =  - {{6x{{\left( {{x^2} + {y^2} - 1} \right)}^2} - 2x{y^3}} \over {6y{{\left( {{x^2} + {y^2} - 1} \right)}^2} - 3\;{x^2}{y^2}}}\tag{2}$$
but from Eq.$(1)$ we can observe that
$${({x^2} + {y^2} - 1)^2} = {x^{{4 \over 3}}}{y^2}\tag{3}$$
Now combining $(2)$ and $(3)$ leads to
$$\eqalign{
  & {{dy} \over {dx}} =  - {{6x\left( {{x^{{4 \over 3}}}{y^2}} \right) - 2x{y^3}} \over {6y\left( {{x^{{4 \over 3}}}{y^2}} \right) - 3\;{x^2}{y^2}}} =  - {{6{x^{{7 \over 3}}}{y^2} - 2x{y^3}} \over {6{x^{{4 \over 3}}}{y^3} - 3{x^2}{y^2}}}  \cr 
  & \,\,\,\,\,\,\,\,\, =  - {{6{x^{{7 \over 3}}} - 2xy} \over {6{x^{{4 \over 3}}}y - 3{x^2}}} =  - {{6{x^{{4 \over 3}}} - 2y} \over {6{x^{{1 \over 3}}}y - 3x}} \cr}\tag{4}$$
and in summary
$${{dy} \over {dx}} =  - {{6x\root 3 \of x  - 2y} \over {6\root 3 \of x y - 3x}}\tag{5}$$
Finally, we can use Eq.$(5)$ to conclude that the slope at $\left( { \pm 1,0} \right)$ is $ \pm 2$ and also that the slope at $\left( {0, \pm 1} \right)$ goes to $ \pm \infty $ depending on whether we are approaching the point from the left or right.  
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