Find the tanget line the curve

Question

2x^3+y^3=6xy at (-1,1) find the tangent line to the curve

Answer 1

Implicit differentiation says:

6x^2 + 3y^2 y' = 6 [ xy' + y]

6x^2 + 3y^2 y' = 6xy' + 6y

6x^2 - 6y = 6xy' - 3y^2 y'

6x^2 - 6y = y' ( 6x - 3y^2)

(6x^2 - 6y) / (6x - 3y^2) = y'

(2x^2 - 2y) / (2x - y^2) = y'

At (-1,1) the slope is M = (2 - 2)/(-2 - 1) = 0

So the horizontal tangent line at (-1,1) is just y=1