x^3 + y^3 =6xy
No need for d/dx just dy/dx for when you find the derivative of y only.
y'= 3x^2+3y^2(dy/dx)= 6y+6x(dy/dx)
dy/dy = (3x^2-6y) /(6x-3y^2)
= (x^2-2y) / (2x-y^2)
y'' = [ (2x- 2(dy/dx)) (2x-y^2) ] - [(x^2-2y)(2-2y(dy/dx))]
= [ 2x-2(x^2-2y/2x-y^2) ] - [(x^2-2y)(2-2y(x^2-2y/2x-y^2)) ] / (2x-y^2)^2
If your having trouble understanding, when i find the derivative of y, its the derivative and dy/dx, where i replace dy/dx in y'' with the dy/dx from y' .
I dont know how to actually put symbols i cant find any button to place actually division signs and such so i clear this up a bit
So then i simplified the complex fraction by multiplying by 2x-y^2
[ 2x(2x-y^2)-2(x^2-2y) ] - (x^2-2y)(2x-y^2) (2(2x-y^2)) (2y(x^2-2y))
4x^2-2y^2 -2x^2-4y - 4y(x^2-2y)^2(2x-y^2)^2
2x^2-2xy^2-4y -4y(x^2-2y)^2(2x-y^2)^2 / (2x-y^2)^2
Is any of this right, that final shown above, not final but mid way, seems way too complicated and much work for such a simple question. All that factoring and such seems alot.
If clarification is needed dont hesitate