I was working with something and have the term ${dy\over dx}\times {dx\over dy}$. In my equality, it must to be $1$, so I try to prove it using the chain rule as follow:
$${dx\over dx} = 1$$ $${dx\over dx} = {d\over dx}(f^{-1}(\ f(x)\ )\ ) = {df^{-1}(f(x))\over df(x)}\times {df(x)\over dx}$$
If we have that $y = f(x)$, so $x = f^{-1}(y)$, then we substitute in the last formula:
$${df^{-1}(f(x))\over df(x)}\times {df\over dx} = {df^{-1}(y)\over dy}\times {dy\over dx}$$ $$ = {dx\over dy}\times {dy\over dx} = 1$$
I should say also,that I was working with functions that are monotonic for all their domain. Is this proof correct?
In the simplest case: Assume you have a diffeomorphism $f: A\to B$ where $A$ and $B$ are open subsets in $\textbf{R}^n$. Then by the chain rule for $x\in A$ and $f(x)=y\in B$:
$$ D(f \circ f^{-1})(y) = (Df)(x) \ \cdot \ D(f^{-1})(y), $$ the dot is here the matrix multiplication and and $Df$ is the Jacobi matrix of the function $f$, which is a matrix valued function on $A$.
Since $D(f \circ f^{-1}) = D(\textrm{id})=1_{n\times n}$ is the constant, you can see that
$$ (Df)(x) \ \cdot \ D(f^{-1})(y) = 1_{n\times n}. $$
In 1D, this reads as
$$ f'(x) \cdot (f^{-1})'(f(x)) = 1 $$
or in the Leibnitz notation (which is far less specific)
$$ \frac{\textrm{d}y}{\textrm{d}x}\frac{\textrm{d}x}{\textrm{d}y}=1. $$