The point(s) on the curve y3+3x2=12y where the tangent is vertical is (are)

Differentiating the given curve w.r.t.  x, we get3y2dydx + 6x = 12dydx ⇒  dydx  =   − 2xy2−4  At point where the tangent(s) is (are) vertical, dydx is not defined,  i.e. at those points.y2−4=0⇒y=±2 When y=2,8+3x2=12(2)⇒3x2=16⇒x=±43 when y=−2,−8+3x2=−24⇒3x2=−16. This is not possible Thus, the required points are (±4/3,2)