Calculus...find d^2y/dx^2: xy+x^3=4.?

My textbook does not provide a single example on how to do this problem. Help por favor.

There are two ways

1) solve for y
xy = 4 - x^3
y = 4/x - x^2
y = (4*x^-1) - (x^2)
y' = (-4*x^-2) - (2x)
y '' = (8x^-3) - 2 = (8/x^3) - (2)

2) implicit derivative

xy+x^3=4

f(x) = x; f'(x) = 1; g(x) = y; g'(y) = y'

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

solve for y'

x*y' = -3*x^2 - y

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

y' = -3x - y/x

f(x) = y; f'(x) = y'; g(x) = x; g'(x) = 1

y'' = (-3) - [(y'*x - y)/x^2]

y'' = [(y - y'*x)/x^2] - (3)

y'' = [(y - (-3x - y/x)*x)/x^2] - (3)

y'' = [(y + (3x + y/x)*x)/x^2] - (3)

y'' = [(y + (3x^2 + y)/x^2] - (3)

y'' = [(2y + 3*x^2)/x^2] - 3

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

y'' = 2y/x^2

y'' = 2*(4/x - x^2)/x^2

y'' = (8/x^3) - (2)

this is Newton's notation rather than Leibniz notation.
d^2/dx^2 simply means the second derivative, or the derivative of the derivative.

dy/dx is: y+3x^2=0

d^2y/dx^2 is 6x