#### Answer

$\frac{d^2y}{dx^2} = \frac{-2y^2-4x^2}{y^3}$

#### Work Step by Step

$2x^2+y^2=4$
First Derivative:
$4x+2y\frac{dy}{dx}=0$
$2y\frac{dy}{dx} = -4x$
$\frac{dy}{dx} = \frac{-2x}{y}$
Second Derivative:
$\frac{d^2y}{dx^2} = \frac{(-2)(y)-(-2x)(\frac{dy}{dx})}{y^2} = \frac{-2y-\frac{4x^2}{y}}{y^2} = \frac{-2y^2-4x^2}{y^3}$