Question 457965
Suppose we take d/dx of both sides:


*[tex \LARGE \frac{d}{dx}(x^2 - 2xy + y^3) = \frac{d}{dx} (1)]


Applying the basic derivation rules such as the power rule, chain rule, and product rule,


*[tex \LARGE 2x - 2(y + x\frac{dy}{dx}) + 3y^2\frac{dy}{dx} = 0]


Solving for dy/dx yields

*[tex \LARGE \frac{dy}{dx} = \frac{2y - 2x}{3y^2 - 2x}]


The slope is vertical when the denominator of dy/dx is zero and the numerator is non-zero. It must be noted that all (x,y) that satisfy 3y^2 - 2x = 0 must also satisfy the original curve.


I did not want to attempt to solve the system of two equations (it would be quite difficult as well since it involves cubics), so I used Wolfram to find the points on the graph where dy/dx has denominator zero. WolframAlpha lists (0.653, -0.660) and (2.046, 1.168) as the only such points. The graph of the curve and the parabola 3y^2 - 2x = 0 can be found by going to the URL below (the intersection of the curves show the desired points):


http://www.wolframalpha.com/input/?i=x^2+-+2xy+%2B+y^3+%3D+1+and+3y^2+-+2x+%3D+0


FYI: Another alternative might be to use polar coordinates and then differentiate. However, this solution might be longer since you will have to solve for r first.