find the points on a curve where the tangent line is horizonta 
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.
 A: So, $2Y=3X^2$
Putting the value of $Y$ in the given equation, $$X^3(27X^3-16)=0\implies X=0,$$ or $X^3=\frac{16}{27}$
But, $X=0\implies Y=0,$ from $2Y=3X^2$
$ \frac{2Y-3X^2}{3Y^2-2X}$ will be $\frac00$ hence undefined.
So, $X\ne0$
$X^3=\frac{16}{27}\implies \frac{3Y^2}{2X}=\frac{3\left(\frac{3X^2}2\right)^2}{2X}=\frac{27}{8}\cdot X^3$ (as $x\ne0$)
$\implies \frac{3Y^2}{2X}=\frac{27}{8}\cdot \frac{16}{27}\ne1\implies 3Y^2-X\ne0$
So, $ \frac{2Y-3X^2}{3Y^2-2X}$ will be $0$ as $2Y=3X^2$ but $3Y^2-2X\ne0$
A: [just to return to this concerning a curious feature of this curve]
The curve in question is the "folium of Descartes", so called as Descartes sent this curve to Fermat as a challenge to the latter's claim for a method of determining slopes of tangent lines.  Fermat handled it successfully, something made a bit more impressive by the fact that he accomplished it in 1638, prior to the more formal development of (infinitesimal) calculus by Newton and Leibniz.  It looks like this:

I have marked horizontal tangent lines in green and vertical ones in red.  I will not reiterate the work already discussed by lab bhattacharjee , but I did want to say something further about the point I raised in the comments.
We find the expression for the first derivative of the implicit functions described by the curve to be
$$ y \ ' \ = \ \frac{3x^2 \ -  \ 2y}{2x \ - \ 3y^2} \ \ . $$
(There is an error in earlier appearances of this rational function, some now corrected.)
As we've already seen this produces horizontal tangents where $ \ y \ = \ \frac{3}{2} x^2 \ $ .  The complication arises when we look for vertical tangents, which occur where $ \ y \ ' \ $ is undefined, that being $ \ x \ = \ \frac{3}{2} y^2 \ $ . (This similarity of the two equations is due to the symmetry of the folium about the line $ \ y \ = \ x \ $ . )  Putting these two equations together gives us two solutions:  $ \ ( \ 0,0 \ ) \ $ and $ \ ( \ \frac{2}{3} , \frac{2}{3} \ ) . $  Only the first of these, however, corresponds to a point on the curve (the latter is not a solution to the equation for the folium).
So the origin is a point which has both a horizontal and a vertical tangent.  This is a situation which can arise for self-intersecting (non-simple) curves.  Finding an "indeterminate" value for $ \ y \ ' \ $ at a point is a sign that the curve has such a self-intersection.
