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.

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 $3Y^2-2X\ne0$
