Answer
$\frac{d^2x}{dy^2} = -\frac{-1}{4y^3}$
Work Step by Step
$x+y^2=1$
Take Derivative and Solve for $\frac{dy}{dx}$
$1+2y\frac{dy}{dx}=0$
$\frac{dy}{dx} = \frac{-1}{2y}$
Take Second Derivative and Solve for $\frac{d^2y}{dx^2}$
$\frac{d^2x}{dy^2} = \frac{2}{4y^2}\times \frac{-1}{2y} = -\frac{-1}{4y^3}$