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Find the points on the curve $x^3 + y^3 = 2xy$ where the line tangent to the curve will be horizontal

I know that that this means that the derivative of the curve will be equal to 0. This is what I get: $$\frac{(2Y-3X^2)}{(3Y^2-X)} = 0$$ ..and then I'm stuck. Please help.

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    Just to add a clarification to lab bhattacharjee's answer, a fraction is equal to 0 when the numerator is equal to 0. So you get $2Y-3X^2=0$ and then ultimately $2Y=3X^2$ (the first line of his answer).2012-12-13
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    that's clear....but what would be the point (s)?, the prof. hinted that the point(s) on the curve most satisfy the equation for the curve x3+y3=2xy2012-12-13
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    Keep in mind that this curve is _not_ the graph of a function: some values of $x$ produce _more than one_ value of $y$. The point $(0,0)$ is on the curve, but makes both the numerator and the denominator in the derivative of the _implicit_ function equal zero. [Hint: on such a curve, sometimes a point can have _two_ slopes!]2013-04-28
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    On reviewing this problem, there is a (small) error in the expression for the slope of the tangent line that will affect the numerical result. This does not alter certain general conclusions, however.2014-05-11

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