Finding $\frac{d^2x}{dy^2}$ in terms of $\frac{d^2y}{dx^2}$ 
Finding $\frac{d^2x}{dy^2}$ in terms of $\frac{d^2y}{dx^2}$

My Attempt:
$\frac{d^2x}{dy^2}=\frac{d}{dy}(\frac{dx}{dy})=\frac d{dx}(\frac{dx}{dy})\frac{dx}{dy}=\frac d{dx}(\frac{dy}{dx})^{-1}\frac{dx}{dy}=-(\frac{dy}{dx})^{-2}\frac{d^2y}{dx^2}\frac{dx}{dy}$
My doubt is in the step: $\frac{d}{dy}(\frac{dx}{dy})=\frac d{dx}(\frac{dx}{dy})\frac{dx}{dy}$
It's like I have broken the operator $\frac d{dy}$ into $\frac d{dx}\times\frac{dx}{dy}$. Is this valid? What can be the explanation for this?
 A: $\newcommand{\d}{\mathrm{d}}$If $y(x)$ admits an inverse $x(y)$ satisfying suitable conditions then $x'(t)=\frac{1}{y'(x(t))}$ from the chain rule applied to $y(x(t))=t$. I use a different letter $t$ for clarity. Then (derivative of a reciprocal): $$x''(t)=-\frac{y''(x(t))\cdot x'(t)}{(y'(x(t)))^2}=-\frac{y''(x(t))}{(y'(x(t)))^3}$$So in loose notation with $t=y$: $$\frac{\d^2 x}{\d y^2}=-\frac{\frac{\d^2y}{\d x^2}}{\left(\frac{\d y}{\d x}\right)^3}$$
An example with $y=x^2$ on the interval $\Bbb R_{\ge0}$:

$\frac{\d y}{\d x}=2x$ and $\frac{\d^2 y}{\d x^2}=2$. The claim is then that: $$\frac{\d^2x}{\d y^2}=-\frac{2}{(2x)^3}=-\frac{1}{4x^3}$$We can check that $x=\sqrt{y}$ gives: $$\frac{\d^2x}{\d y^2}=\frac{\d x}{\d y}\left[\frac{1}{2\sqrt{y}}\right]=-\frac{1}{4y\sqrt{y}}=-\frac{1}{4x^3}$$As claimed.

Another example with $y=\sin(x)$ on $[0,\pi/2]$:

$\frac{\d y}{\d x}=\cos(x)$ and $\frac{\d^2y}{\d x^2}=-\sin(x)$. The claim is then that: $$\frac{\d^2x}{\d y^2}=-\frac{-\sin(x)}{\cos^3(x)}=\frac{\sin(x)}{\cos^3(x)}$$We can check that $x=\arcsin(y)$ (principal branch) gives: $$\begin{align}\frac{\d^2x}{\d y^2}&=\frac{\d x}{\d y}\left[\frac{1}{\sqrt{1-y^2}}\right]\\&=-\frac{1}{2}\frac{-2y}{(1-y^2)\sqrt{1-y^2}}\\&=\frac{y}{(1-y^2)\sqrt{1-y^2}}\\&=\frac{\sin(x)}{\cos^3(x)}\end{align}$$As claimed.

With regards to your solution, it is difficult to parse. The Leibniz notation is what is used most often lower down the education system because it is easy to palm off as "fractions", but there is real danger in its overusage (you can never be sure if you did something because: "fractions!" or because your manipulation was valid ;)). Being careful with the functions at play will always be a safe approach.
