When is $(dy/dx)^{-1}=dx/dy$ for multivariable functions? In general for a multivariable functions $y(m,n)$, $x(m,n)$,
$$\frac{dy}{dx}=\frac{\partial y}{\partial m}\frac{dm}{dx}+\frac{\partial y}{\partial n}\frac{dn}{dx}\tag{1}$$
and
$$\frac{dx}{dy}=\frac{\partial x}{\partial m}\frac{dm}{dy}+\frac{\partial x}{\partial n}\frac{dn}{dy}\tag{2}$$
so in general clearly $(dy/dx)^{-1}$ need not be equal to $dx/dy$.
My question is what is the precision Mathematical conditions on functions $y(m,n)$ and $x(m,n)$ under which this equality holds.
Clearly I could explicitly calculate $dy/dx$ and $dx/dy$ and check the equality, but is there a better way to see this equality from the point of view of the properties of the functions. Here, by functions I mean functions that commonly appear in physics (and definitely not some exotic function that has no physical origins - this is the reason I decided to post it in PhysicsSE and not MathSE).
If there exist any such properties, I would also like to have a mathematical proof that states that these properties are equivalent to proving the equality explicitly from the first principle i.e. using (1) and (2).

Edit: I have been notified of a possible ill definition of these particular chain rules, which would mean the question itself is invalid and I very likely might have misinterpreted something. To verify this possible ill definition I have posed a question in MathSE.
 A: Its only true when $y$ is a single variable function of $x$, so
$$
y = y(x).
$$
If $y$ is a function of multiple variables, say $y = y(m,n)$, then its derivative is really a vector.
$$
\nabla y = \left( \frac{\partial y}{\partial m}, \frac{\partial y}{\partial n}\right).
$$
Furthermore, the derivative of $x$ is also a vector, $\nabla x$.
Now, note that you can't even divide vectors even if you wanted to. In general, the expression
$$
\frac{\nabla y}{\nabla x}
$$
just makes no sense. So then, in the multivariable case, the expression
$$
\frac{dy}{dx}
$$
likewise makes no sense in general.
However, there is a special case in which it does make sense. That is when $y$ really can be written as a function of $x$, i.e.
$$
y(m,n) = f( x(m, n) )
$$
for some single variable function $f$. In this special case, $\nabla y$ is actually parallel to $\nabla x$, so it sort of does make sense to take the quotient $\nabla y / \nabla x$. And also, because $y = f(x)$, $y$ really is in a way just a single variable function of $x$, and so your identity will hold only in this case.
Also, as written now, your equations $(1)$ and $(2)$ are simply incorrect. $m$ and $n$ are not functions of $x$, so the expression $\partial m / \partial x$ doesn't make much sense. It's an incorrect application of the chain rule. Therefore you won't be able to prove anything from those starting points. To reiterate, the expression $dy/dx$ just has no meaning when $y$ and $x$ are general functions of $m$ and $n$, simply because a value of $x$ does not determine a unique value of $y$.
Edit:
As for your equations $(1)$ and $(2)$, here is the "meaning" of those equations. Each pair of functions of two variables defines a Jacobian which is a $2 \times 2$ matrix. So for the functions $(y(m,n), x(m,n))$ we have the Jacobian
$$
J_{x(m,n), y(m,n)} = \begin{pmatrix} \tfrac{\partial x}{\partial m} & \tfrac{\partial x}{\partial n} \\ \tfrac{\partial y}{\partial m} & \tfrac{\partial y}{\partial n} \end{pmatrix}
$$
and for the functions $(m(x,y), n(x,y))$ we have the Jacobian
$$
J_{m(x,y), n(x,y)} = \begin{pmatrix} \tfrac{\partial m}{\partial x} & \tfrac{\partial m}{\partial y} \\ \tfrac{\partial n}{\partial x} & \tfrac{\partial n}{\partial y} \end{pmatrix}.
$$
Because the maps here are inverse functions, one can prove that $J_{x(m,n), y(m,n)} J_{m(x,y), n(x,y)}= I$.
If we want to calculate the Jacobian for the composition of the two maps, which is $(x(m(x,y), n(x,y)), y(m(x,y), n(x,y) )$, the correct formula to use is to multiply the two Jacobians together. However, this is just $I$, which tautologically confirms that
$$
\begin{pmatrix} \tfrac{\partial x}{\partial x} & \tfrac{\partial x}{\partial y} \\ \tfrac{\partial y}{\partial x} & \tfrac{\partial y}{\partial y} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.
$$
However, you won't be able to draw any conclusions about the form of $x$ or $y$ from this calculation. All you can do is calculate the identity matrix tautologically.
A: You're allowed to do this when you're working with functions of a single variable.
If you check, you'll find that in all examples in physics where $dy / dx = (dx / dy)^{-1}$ is used, this is the case. Examples include the trajectory of a single particle, which can be parametrized as $x(t)$, and the potential in a problem with spherical symmetry, which can be parametrized as $V(r)$.
A: As @knzhou has noted, in case of a function of a single variable
$$
\left[y'(x)\right]^{-1} = \frac{dx}{dy},
$$
since the derivative and the differential of a function of a single variable are equivalent.
This is not the case for the functions of several variables, see, e.g., my answers here and here.
A: It may not be obvious but you haven't specified all the information yet. Excuse me for changing your notation but define
\begin{align}
x(\vec u):\quad \mathbb R^n\rightarrow\mathbb R\\
y(\vec u):\quad \mathbb R^n\rightarrow\mathbb R
\end{align}
where I redefined $(m,n)$ as an n-dimensional point $\vec u$. Right now when I try to calculate $\frac{dy}{dx}$ I will get stuck because I'm missing information. You are differentiating $y(x)=y(\vec u(x))$ with respect to $x$ but we have never specified $\vec u(x)$. We can't just invert $x(\vec u)\rightarrow\vec u(x)$ because for a given value $x$ there's an infinite number of points $\vec u$ that give the same value $x$. To define a total derivative you implicitly have to specify a path in $\mathbb R^n$ because by definition a total derivative only depends on one variable. Once I have specified $\vec u(x)$ it is obvious that $dy/dx=(dx/dy)^{-1}$ because $y(\vec u(x))$ is just a function of one variable i.e. you can write it as $y(x)$. If $y(x)$ is invertible then
\begin{align}
\frac{dy}{dy}=\frac{d}{dy}y(x(y))&=\frac{dy(x(y))}{dx}\frac{dx(y)}{dy}\\
&=\frac{dy}{dx}\frac{dx}{dy}\overset !=1
\end{align}
So the mathematical conditions would be that $x(\vec u),\vec u(x),y(\vec u),\vec u(y)$ all exist and are continuous. I don't know much about function analysis so maybe there are better conditions.
