please help; calc
posted by stephanie on .
original curve: 2y^3+6(x^2)y12x^2+6y=1
dy/dx=(4x2xy)/(x^2+y^2+1)
a) write an equation of each horizontal tangent line to the curve
b) the line through the origin with the slope .1 is tangent to the curve at P. Find x and y of point P.

if dy/dx is zero, that must mean 4x2xy is zero, or y=2
When y=2, solve for x in the original equation. 
when you plug it in the original eqn. it doesn't work

28=1 is the value of x and that doesn't work