Differentiate two times a simple expression Let's say I have:
$$ y = -\frac{1}{\tan(x)} $$
How do you get a relation between $\mathrm{d}^2x$ and $\mathrm{d}^2y$ ?
What I know:
$$\mathrm{d}y = \frac{\mathrm{d}x}{\sin(x)^2}  $$

$$y=-\frac{1}{\tan(x)}=-\cot(x)$$ and $$\frac{dy}{dx}=\csc^2(x)$$
To find $\frac{\,d^2y}{\,dx^2}$, differentiate $\csc^2(x)$ once more to get $$\frac{\,d^2y}{\,dx^2}=\frac{\,dy}{\,dx}\csc^2(x)=-2\csc^2(x)\cot(x)\Longrightarrow d^2y=-2\csc^2(x)\cot(x)dx^2\tag{*}$$
Then, $$y=-\cot(x)\Longrightarrow x=\cot^{-1}(-y) $$ so now, $$\frac{dx}{dy}=\frac{1}{1+y^2}\Longrightarrow \frac{d^2x}{dy^2}=-\frac{2y}{(1+y^2)^2}\Longrightarrow d^2x=-\frac{2y}{(1+y^2)^2}dy^2\tag{**}$$
From $(*)$, we have
$$d^2y=2\csc^2(x)ydx^2 \Longrightarrow y=\frac{d^2y}{dx^2}\frac{\sin^2(x)}{2}$$
Substitute $y$ into $(**)$ for your relation.
Not really sure  what meaning or significance it has; it seems like an abuse of notation. This would only work on certain constrained domains of $y=-\cot(x)$
