Finding the second derivative by implicit differentiation

[ireallyhatemath]

Find (d^2y)/(dx^2) of x^3 + y^3 = 1

So far I have figured out the first derivative...
3x^2 + 3y^2(dy/dx) = 0
3y^2(dy/dx) = -3x^2
(dy/dx) = -3x^2/3y^2
(dy/dx) = -x^2/y^2

But then what do i do? Would I use the quotient rule and then plug in for y prime to find the second derivative?

 

[Unknown008]

Find (d^2y)/(dx^2) of x^3 + y^3 = 1

So far I have figured out the first derivative...
3x^2 + 3y^2(dy/dx) = 0
3y^2(dy/dx) = -3x^2
(dy/dx) = -3x^2/3y^2
(dy/dx) = -x^2/y^2

But then what do i do? Would I use the quotient rule and then plug in for y prime to find the second derivative?

Yes, can you post what you get?

 

[galactus]

Would I use the quotient rule and then plug in for y prime to find the second derivative?

Yes.