- #1

- 16

- 0

## Homework Statement

Find the coordinates of the stationary points on the curve:

x^3 + (3x^2)(y) -2y^3=16

## Homework Equations

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 :)