Given curve is y3 – 3xy + 2 = 0
Since the tangent is horizontal, so dy/dx = 0
⇒ y = 0
which is not possible, since the point y = 0 does not lie on the curve.
Thus, H = ϕ = null set
When the tangent is vertical,so dx/dy = 0
⇒ (y2 – x)/y = 0
⇒ x = y2
when x = y2, then y3 – 3y3 + 2 = 0
⇒ 2y3 = 2
⇒ y3 = 1
⇒ y = 1 when y = 1, then x = 1
Hence, V = {(1, 1)}