Posted by jahil on Sunday, November 21, 2010 at 6:46pm.
x^3+3x^2y+y^3=8
3x^2 + 3x^2 dy/dx + y(6x) + 3y^2 dy/dx = 0
dy/dx(3x^2 + 3y^2) = -3x^2 - 6xy
dy/dx = -3x(x + 2y)/(3(x^2 + y^2)
= -x(x+2y)/(x^2 + y^2)
or
= -(x^2 + 2xy)/(x2 + y^2)
I don't understand the answer the way you typed it, but there should definitely be a negative in there.