[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.