how to find equation of horizontal tangent line of 2y^3+6x^2y-12x^2+6y=1

To find the equation of the horizontal tangent line of the given curve, follow these steps:

Step 1: Differentiate the equation with respect to x to find the derivative.
dy/dx = (-6x^2 - 12xy + 12x) / (6y^2 + 6)

Step 2: To find the points on the curve where the slope of the tangent line is zero (i.e., horizontal), set the derivative equal to zero and solve for x.
(-6x^2 - 12xy + 12x) / (6y^2 + 6) = 0

Step 3: Simplify the equation:
-6x^2 - 12xy + 12x = 0

Step 4: Factor out common terms:
-6x(x + 2y - 2) = 0

Step 5: Set each factor equal to zero and solve for x:
-6x = 0 -> x = 0
x + 2y - 2 = 0 -> x = 2 - 2y

Step 6: Substitute the values of x back into the original equation to find the corresponding y-values.
For x = 0:
2y^3 + 0 - 12(0)^2 + 6y = 1
2y^3 + 6y = 1
2y(y^2 + 3) = 1
y(y^2 + 3) = 1/2

For x = 2 - 2y:
2y^3 + 6(2 - 2y)^2y - 12(2 - 2y)^2 + 6y = 1
Simplify and solve the equation for y.

Step 7: Substitute the values of x and y into the equation of a line, y = mx + b, to find the equation of the horizontal tangent line.

By following these steps, you should be able to find the equation of the horizontal tangent line for the given curve.