Derive each term with respect to itself

x^3 + y^3 - 3x - 6 = 0

3x^2 * dx + 3y^2 * dy - 3 * dx - 0 = 0

3x^2 * dx + 3y^2 * dy - 3 * dx = 0

x^2 * dx + y^2 * dy - 3 * dx = 0

Group by the differential

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

dy * y^2 = dx * (3 - x^2)

Solve for dy/dx

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

Turning points occur when dy/dx = 0

3 - x^2 = 0

3 = x^2

x = -sqrt(3) , sqrt(3)

Plug that into the original equation and solve for y

(-sqrt(3))^3 + y^3 - 3 * (-sqrt(3)) - 6 = 0

-3 * sqrt(3) + 3 * sqrt(3) + y^3 = 6

y^3 = 6

y = 6^(1/3)

sqrt(3)^3 + y^3 - 3 * sqrt(3) - 6 = 0

3 * sqrt(3) - 3 * sqrt(3) + y^3 = 6

y^3 = 6

y = 6^(1/3)

The turning points are at:

(-(3^(1/2)) , 6^(1/3))

(3^(1/2) , 6^(1/3))

To differentiate y^3 with respect to x, use the chain rule,

d(y^3)/dx= (dy^3/dy)(dy/dx = 3y^2dy/dx.

Differentiate each term of the equation with respect to x and you get

3x^2+3y^2dy/dx -3=0 divide by 3 to give x^2+y^2dy/dx-1=0

so y^2dy/dx = 1-x^2 and dy/dx=(1-x^2)/y^2

The turning points are where dy/dx=0 giving 1-x^2=0 so x^2=1

and x=-1, and sub into the given equation to get -1+y^3+3-6=0

giving y^3=4 ,y=cube root of 4

or x=1, y^3=8, y=2

The two turning points have coordinates (-1, 4^(1/3))and (1, 2)

3x² + 3y² dy/dx - 3 = 0

y² dy/dx = 1 - x²

dy/dx = (1 - x²) / y² = 0 for turning points

x = ± 1

1 + y³ - 3 - 6 = 0

y³ = 8

y = 2

Thus (1,2) is a turning point

-1 + y³ + 3 - 6 = 0

y³ = 4

y = 4^(1/3)

Thus (-1, 4^(1/3) ) is a turning point