HOW Do YOU Solve DERIVATIVES Through A Point

Find the slope of the folium of Descartes x^3+y^3-9xy=0 at the points (4,2) and (2,4).

Slope at (4,2)=
Slope at (2,4)=

At what point other than the origin does the folium have a horizontal tangent?

The point (x,y)=()




thanks

x^3 + y^3 - 9xy = 0
3x^2 * dx + 3y^2 * dy - 9x * dy - 9y * dx = 0
dx * (3x^2 - 9y) + dy * (3y^2 - 9x) = 0
dx * 3 * (x^2 - 3y) + dy * 3 * (y^2 - 3x) = 0
dx * (x^2 - 3y) + dy * (y^2 - 3x) = 0
dx * (x^2 - 3y) = dy * (3x - y^2)
dy/dx = (x^2 - 3y) / (3x - y^2)


(4,2)

dy/dx = (4^2 - 3 * 2) / (3 * 4 - 2^2)
dy/dx = (16 - 6) / (12 - 4)
dy/dx = 10/8
dy/dx = 5/4

(2,4)

dy/dx = (2^2 - 3 * 4) / (3 * 2 - 4^2)
dy/dx = (4 - 12) / (6 - 16)
dy/dx = -8/-10
dy/dx = 4/5



dy/dx = (x^2 - 3y) / (3x - y^2)
dy/dx = 0

x^2 - 3y = 0
x^2 = 3y
y = (1/3) * x^2


x^3 + y^3 - 9xy = 0
x^3 + (1/3)^3 * (x^2)^3 - 9 * x * (1/3) * x^2 = 0
x^3 + (1/27) * (x^6) - 3x^3 = 0
-2x^3 + (1/27) * x^6 = 0
(1/27) * x^6 = 2x^3
(1/27) * x^3 = 2
x^3 = 2 * 27
x = 3 * (2^(1/3))

y = (1/3) * x^2
y = (1/3) * 9 * 2^(2/3)
y = 3 * 2^(2/3)

(3 * 2^(1/3) , 3 * 2^(2/3))