\(\frac{dx}{dy} = \frac{1}{y'}\), where \(y' = \frac{dy}{dx}\).

\(\frac{d^2x}{dy^2} = \frac{d}{dy}\left(\frac{1}{y'}\right) = \frac{d}{dx}\left(\frac{1}{y'}\right) \cdot \frac{dx}{dy} = -\frac{y''}{(y')^2} \cdot \frac{1}{y'} = -\frac{y''}{(y')^3}\)

So \(\frac{d^2x}{dy^2} = -\left(\frac{d^2y}{dx^2}\right)\left(\frac{dy}{dx}\right)^{-3}\).

Answer: (D)