Show that $\frac{d^2y}{dx^2}=-\frac{d^2x}{dy^2} / \left(\frac{dx}{dy}\right)^3$ is an identity From Applied Differential Equations (first edition) by Spiegel,

Show that $$\frac{d^2y}{dx^2} = -\frac{d^2x}{dy^2}/\left(\frac{dx}{dy}\right)^3$$ is an identity.
  Hint: Differentiate both sides of $\frac{dy}{dx} = \left(\frac{dx}{dy}\right)^{-1}$ with resepct to $x$.

After looking at the hint, I got $$\frac{d^2y}{dx^2} = -\left(\frac{dx}{dy}\right)^{-2}\frac{d}{dx}\left(\frac{dx}{dy}\right)$$ but I'm not sure how to proceed from here. Using $dx/dy = (dy/dx)^{-1}$ doesn't help, but otherwise I can't think of anyway to evaluate $\frac{d}{dx}\left(\frac{dx}{dy}\right)$.

From the hint, $$\frac{dy}{dx} = \left(\frac{dx}{dy}\right)^{-1}$$ Differentiate both sides with respect to $x$ to obtain $$\frac{d^2y}{dx^2} = -\left(\frac{dx}{dy}\right)^{-2}\frac{d}{dx}\left(\frac{dx}{dy}\right)$$ Thanks to Fabien's prodding, remember that $\frac{dx}{dy}$ is a function of $y$, and apply the chain rule and the hint again to get $$\frac{d}{dx}\left(\frac{dx}{dy}\right) = \frac{d^2x}{dy^2}\frac{dy}{dx} = \frac{d^2x}{dy^2}\left(\frac{dx}{dy}\right)^{-1}$$ Substituting this yields $$\frac{d^2y}{dx^2} = \frac{d^2x}{dy^2}/\left(\frac{dx}{dy}\right)^3$$
