Find the coordinates of the stationary points on the curve:
x^3 + (3x^2)(y) -2y^3=16
Stationary points occur when the first derivative of y with respect to x is equal to zero
The Attempt at a Solution
I implicitly differentiated the equation and got
dy/dx = (x^2 + 2xy) / (2y^2 - x^2)
I know I have to make this equal to zero but then I'm not sure how to find the x and y coordinates of the stationary point.
Help would be greatly appreciated :)