i need more help with another AP calculus problem?

Question

1. f(x)= tan(3^x) has an interval in [0, 1.4] what's the derivative at this point.
2. consider the curve: 2y^3+6x^2y-12x^2=6y=1
dy/dx=(4x-2xy)/(x^2+y^2+1)
...write an equation of each horizontal tangent line to the cruve.
...the line through the origin with a slope of -1 is tangent to the curve at point p. find the x- and y-coordinate of point p.

Answer 1

f´(x)= (1+tan^2(3^x))* 3^x*ln3
At x= 1.4 f¨=28.926
2)6y^2y´+12xy+6x^2y´-24x+6y´=0
y´(y^2+x^2+1)= 4x-2xy
y´=0 so 4x-2xy=0 so x(4-2y)=0
x=0 2y^3+6y-1=0 root =0.1652 equation y=0.1652
y=2 so 16 +12x^2-12x^2+12=1 (No point)
3)y=-x is the tangent
2y^3+6y^3-12y^2+6y=1
8y^3-12y^2+6y-1=0 Factorize as (2y-1)^3 =0 so the point is(-1/2,1/2)