Find y" by implicit differentiation

x^3 + y^3 = 1

Here's my work:

3x^2 + 3y^2y' = 0

y' = (-3x^2)/(3y^2) = (-x^2)/(y^2)

y" = (-2xy^2 + 2x^2y(dy/dx)) / y^4

= 2(-xy - x^4y^-2) / y^3

The back of the book says that the answer is -2x / y^5. I don't understand how they're simplifying it to that. Thanks for any help.