Prove $\frac{d^2y}{dx^2}=-\frac{\frac{d^2x}{dy^2}}{(\frac{dx}{dy})^3}$ 
Prove $$\dfrac{d^2y}{dx^2}=-\dfrac{\dfrac{d^2x}{dy^2}}{\left(\dfrac{dx}{dy}\right)^3}$$

Tried
$$\dfrac{dy}{dx}=\dfrac{1}{\dfrac{dx}{dy}}$$
So
$$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)=\dfrac{-\dfrac{d}{dx}\left(\dfrac{dx}{dy}\right)}{\left(\dfrac{dx}{dy}\right)^2}$$
But cannot go any further, any help? Thanks~
 A: Here is a solution using functional notations.
Let $f:x \mapsto y$ and $g:=f^{-1}:y \mapsto x.$
Let us consider a certain point $x_0$ and its image $y_0$ by $f$.
We assume there is a certain neigborhood $U$ of $x_0$ and $V$ of $y_0$ where (the restriction of) $f$ is a bijection. Moreover, we assume that, in these neighborhoods, $f$ and $g$ are $C^2$.
Differentiating relationship
$$\tag{1}f(g(u))=u.$$
(we have taken $u$ instead of $x$ or $y$ in order to have a "neutral" variable).
we have :
$$\tag{2}f'(g(u))g'(u)=1.$$
Assuming $g'(u) \neq 0$, (2) is equivalent to:
$$\tag{3}f'(g(u))=\frac{1}{g'(u)}.$$
Differentiating (2), one gets:
$$\tag{4}f''(g(u))(g'(u))^2+f'(g(u))g''(u)=0.$$
Otherwise said:
$$\tag{5}f''(g(u))=-\dfrac{f'(g(u))}{(g'(u))^2}g''(u).$$
Taking (3) into account in (5), we obtain the desired result under the form: 
$$\tag{6}f''(g(u))=-\dfrac{1}{(g'(u))^3}g''(u).$$
Remark: I just noticed a reference:(http://elib.mi.sanu.ac.rs/files/journals/tm/20/tm1112.pdf) where this property is proven in the same way.
A: Setting
$$
z=\frac{dx}{dy},
$$
we have
$$
\frac{d}{dx}=\frac{dy}{dx}\frac{d}{dy}=z^{-1}\frac{d}{dy}.
$$
It follows that
\begin{eqnarray}
\frac{d^2y}{dx^2}&=&\frac{d}{dx}(z^{-1})=z^{-1}\frac{d}{dy}(z^{-1})=z^{-1}\left(-z^{-2}\frac{dz}{dy}\right)=-z^{-3}\frac{dz}{dy}=-\dfrac{\dfrac{dz}{dy}}{z^3}
\end{eqnarray}
Hence
$$
\frac{d^2y}{dx^2}=-\dfrac{\dfrac{d^2x}{dy^2}}{\left(\dfrac{dx}{dy}\right)^3}
$$
