Validity of this particular "Chain Rule" For two multivariable function $y(m,n)$ and $x(m,n)$ does it make sense/is it mathematically correct to talk of the following chain rules -
$$\frac{dy}{dx}=\frac{\partial y}{\partial m}\frac{dm}{dx}+\frac{\partial y}{\partial n}\frac{dn}{dx}\tag{1}$$
and
$$\frac{dx}{dy}=\frac{\partial x}{\partial m}\frac{dm}{dy}+\frac{\partial x}{\partial n}\frac{dn}{dy}\tag{2}$$
simultaneously, where m, n can be rewritten as $m=m(x,y),\,n=n(x,y)$ and $d/dx,\,d/dy$ of m, n exists.
The context is this PhysicsSE question - if that is relevant.
How correct/wrong is this "Chain rule"?
Any intuitive/geometric argument together with a mathematical proof, if possible, is highly appreciated.
 A: The expression $\frac{dy}{dx}$ suggests that we have a rule which associates a value of $x$ to a single, well-defined value of $y$ - let's say $y=g(x)$ - and that we are differentiating it, so $\frac{dy}{dx} = g'(x)$.  In your example, you have one function of two variables $f:\mathbb R^2 \rightarrow \mathbb R$ and two functions of one variable $m,n : \mathbb R \rightarrow \mathbb R$; $g$ is then given by their composition,
$$y = g(x) = f\bigg(m(x),n(x)\bigg)$$
$$\implies \frac{dy}{dx}= g'(x) = (\partial_1f)\big(m(x),n(x)\big)\cdot m'(x) + (\partial_2f)\big(m(x),n(x)\big) \cdot n'(x)$$
where $\partial_i f$ is the partial derivative of $f$ with respect to its $i^{th}$ entry.  Standard convention is to suppress the arguments and associate each entry with the function we plug into it, which leads to the prettier expression
$$\frac{dy}{dx} = g'(x) = \frac{\partial f}{\partial m}  \frac{dm}{dx} + \frac{\partial f}{\partial n} \frac{dn}{dx}$$

It is entirely possible that $g$ can be inverted, at least locally.  This would imply the existence of a function $\tilde g$ such that $\tilde g(g(x)) = g(\tilde g(x)) = x$, which means that we can write $x = \tilde g(y)$.  It's not hard to show that $\tilde g'(y) = \frac{1}{g'\big(\tilde g(y)\big)}$ as long as $g'\big(\tilde g(y)\big)\neq 0$; in such cases, we would have that
$$\frac{dx}{dy} = \tilde g'(y) = \frac{1}{g'\big(\tilde g(y)\big)} = \left(\left.\frac{dy}{dx}\right|_{x=\tilde g(y)}\right)^{-1}$$
However, this $\tilde g$ will generally not be able to expressed in terms of the functions $f$, $m$, and $n$ in any meaningful way.  Take for example $f(m,n)=m^2n$, $m(x)=x$, $n(x)=x^4$.  Then $y=g(x) = f\big(m(x),n(x)\big) = x^6$, and $\frac{dy}{dx} = 6x^5$.
Inverting this, $x = \tilde g(y) = y^{1/6}$, so $\frac{dx}{dy} = \frac{1}{6 y^{5/6}} = \left(6 x^5\right)^{-1}$, and all is well.

Your expression (2) implies that $\tilde g(y)$ can be expressed as $F\big(m(y),n(y)\big)$ for yet another function $F:\mathbb R^2\rightarrow \mathbb R$.  In such a case, we would indeed have (skipping right to the pretty notation),
$$\frac{dx}{dy} = \tilde g'(y) = \frac{\partial F}{\partial m}\frac{dm}{dy} + \frac{\partial F}{\partial n}\frac{dn}{dy}$$
In my simple example above, note that $\tilde g(y) = y^{1/6}$, so we could let $F(m,n) = m^{1/6}$.  Of course, this $F$ is not unique; $F(m,n)=n^{1/12}$ and $F(m,n) = m^{1/12} n^{1/24}$ would also work, and would also give us the result we're looking for.
More generally, if $m$ is invertible with inverse $\tilde m$, we could let $F(m,n) = \tilde g\big(\tilde m (m)\big)$, which would yield the result we're looking for.

Of course, the above is what I would consider idle mathematical curiosity.  Given a relationship $y = g(x) = f\big(m(x),n(x)\big)$, it is broadly possible to find a function $F$ such that $x = \tilde g(y) = F\big(m(y),n(y)\big)$. This $F$ is highly non-unique in general, and I can think of no particularly good reason to do this, but you can bend over backward to do so if you really want to.  However, $F$ could not be determined from $f$ alone; it would inevitably involve the inverse function of $m$, $n$, or both.
