Answer
$\frac{d^2y}{dx^2} = \frac{-3x^2y^4-3x^6}{y^7}$
Work Step by Step
$x^4+y^4=64$
First Derivative:
$4x^3+4y^3 \frac{dy}{dx} = 0$
$\frac{dy}{dx} = \frac{-x^3}{y^3}$
Second Derivative:
$\frac{d^2y}{dx^2} = \frac{(-3x^2)(y^3)-(-x^3)(3y^2)(\frac{dy}{dx})}{y^6} = \frac{-3x^2y^4-3x^6}{y^7}$
