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 $\frac{d}{dx}\left(\frac{dx}{dy}\right)$.

With $f(g(y))=y$ you get in the first two derivatives
$$
f'(g(y))g'(y)=1
$$
and
$$
f''(g(y))g'(y)^2+f'(g(y))g''(y)=0
$$
Multiply with $g'(y)$ to obtain
$$
f''(g(y))g'(y)^3=-g''(y)
$$
and with $x=g(y)$, $y=f(x)$
$$
f''(x)=-g'(y)^{-3}g''(y)
$$
