The original equation is $$x^2-2xy+y^3= 4$$ and I hope the derivative to be $$\frac{dy}{dx} = \frac{2(x-y)}{2x-3y^2}$$
I know the vertical tangent is when the denominator is $0$, but I am having trouble determining the vertical tangent.
[Math] Vertical Tangent line with Implicit Differentiation
implicit-differentiation
Related Solutions
As OP already found, the expression for the slope of a tangent line to a point on the curve in question is
$$ y \ ' \ = \ \frac{y^2 \ - \ 3x^2}{3y^2 \ - \ 2xy} \ = \ \frac{y^2 \ - \ 3x^2}{y \ (3y \ - \ 2x)} \ \ . $$
Something useful to check for immediately is whether this rational function can be "indeterminate", that is, whether both numerator and denominator can be zero at the same time for a point on the curve. There are two cases to consider, due to the two factors in the denominator:
I -- $ \ y^2 \ = \ 3x^2 \ \ , \ \ y \ = \ 0 \ \ \Rightarrow \ \ x \ = \ 0 \ $ ; however, the point $ \ ( 0 , 0 ) \ $ does not lie on this curve
II -- $ \ y^2 \ = \ 3x^2 \ \ , \ \ 3y \ = \ 2x \ \ \Rightarrow \ \ \left( \frac{2}{3}x \right)^2 \ = \ 3 x^2 \ \ \Rightarrow \ \ x \ = \ 0 \ \ , \ \ y \ = \ 0 \ $ ; again, this is not a solution to the curve equation.
So the issue of "slope indeterminancy" does not occur for this curve, as it does for one of its "cousins", the folium of Descartes (discussed, for instance, here).
We can then locate the points at which horizontal tangents occur by inserting the relation $ \ y^2 \ = \ 3x^2 \ \Rightarrow \ y \ = \ \pm \ 3^{1/2}x \ $ into the equation for the curve (as mentioned by Graham Kemp), thus:
$$ x^3 \ + \ y^3 \ - \ xy^2 \ = \ 8 \ \ \Rightarrow \ \ x^3 \ + \ \left( \pm \ 3^{1/2}x \right)^3 \ - \ x \ \left( \pm \ 3^{1/2}x \right)^2 \ = \ 8 $$
$$ \Rightarrow \ \ \left( 1 \ \pm \ 3^{3/2} \ - \ 3 \right) \ x^3 \ = \ 8 $$
$$ \Rightarrow \ \ x \ = \ \left[ \frac{8}{-2 \ \pm \ 3 \sqrt{3}} \right]^{1/3} = \ \frac{2}{( -2 \ \pm \ 3 \sqrt{3} \ )^{1/3}} \ \approx \ -1.036 \ , \ 1.358 $$
$$ y \ = \ \pm \ \sqrt{3} \ \cdot \ x \ \rightarrow \ 1.794 \ , \ 2.352 \ \ . $$
[Upon insertion into the original curve equation, we see that the negative values for $ \ y \ $ are not solutions.] Thus, the horizontal tangent lines are found at
$$ \left( \frac{2}{( -2 \ - \ 3 \sqrt{3} \ )^{1/3}} \ , \ \frac{2 \ \sqrt{3}}{( -2 \ - \ 3 \sqrt{3} \ )^{1/3}} \right) \ \ , \ \ \left( \frac{2}{( -2 \ + \ 3 \sqrt{3} \ )^{1/3}} \ , \ \frac{2 \ \sqrt{3}}{( -2 \ + \ 3 \sqrt{3} \ )^{1/3}} \right) \ \ . $$
Vertical tangents occur at the points for which one of the factors in the denominator of the expression for $ \ y \ ' \ $ equals zero:
$$ y \ = \ 0 \ \ \Rightarrow \ \ x^3 \ + \ 0^3 \ - \ x \cdot 0^2 \ = \ 8 \ \ \Rightarrow \ \ x \ = \ 2 \ \ ; $$
$$ y \ = \ \frac{2}{3}x \ \ \Rightarrow \ \ x^3 \ + \ \left( \frac{2}{3}x \right)^3 \ - \ x \cdot \left( \frac{2}{3}x \right)^2 \ = \ 8 $$
$$ \Rightarrow \ \ \left( 1 \ + \ \frac{8}{27} \ - \ \frac{4}{9} \right) \ x^3 \ = \ \left( \frac{27 \ + \ 8 \ - 12}{27} \right) x^3 \ = \ 8 $$
$$ \Rightarrow \ \ x \ = \ \left( \frac{ 8 \ \cdot \ 27}{23} \right)^{1/3} \ = \ \frac{6}{23^{1/3}} \ \approx \ 2.110 \ \ \Rightarrow \ \ y \ = \ \frac{2}{3} \cdot \ \frac{6}{23^{1/3}} \ = \ \frac{4}{23^{1/3}} \ \approx \ 1.407 \ \ . $$
So the points where vertical tangents are found are
$$ ( \ 2, 0 \ ) \ \ \text{and} \ \ \left( \frac{6}{23^{1/3}} , \frac{4}{23^{1/3}} \right) \ \ . $$
A graph of the curve is presented below.
horizontal tangents are marked in green, vertical tangents, in red
Best Answer
The equation $$\bigl(f(x,y):=\bigr)\qquad x^2-2xy+y^3=4$$ defines a set $S$ in the $(x,y)$-plane. This set is, generally speaking, a curve, maybe with singularities. You want to know the points of $S$ where the tangent to $S$ is vertical. Now $S$ can be considered as a level line of the function $f$. Level lines are at each of their points orthogonal to $\nabla f$ at this point. It follows that at the points $p\in S$ where the tangent to $S$ is vertical the gradient $\nabla f(p)$ has to be horizontal, which means that $f_y(x,y)=0$ at such points. Therefore these $p=(x,y)$ will come to the fore by solving the system $$x^2-2xy+y^3=4, \quad -2x+3y^2=0\ .$$ Maybe this system has several solutions $p_k=(x_k,y_k)$. The equation of the tangent to $S$ at $p_k$ is then given by $x=x_k$.