Question regarding differentiation

I have a question on differentiation.

Suppose we have $y=f(x)$.

Is ${dy\over dx}={1 \over {dx \over dy}}$?

I know we can find some counterexamples to it, like in polar coordinates:

$$x=r\cos(\theta),\ y=r\sin(\theta)$$ $${\partial x\over \partial r}=\cos(\theta)={\partial r\over \partial x}$$

This part comes from the Laplace equation in PDE:

Given the following equation:

$\Delta u=0,\ u=u(x,y)$ $Change\ x=r\cos(\theta),\ y=r\sin(\theta)$

The Jacobian of the transformation is

$$\left( \begin{array}{cc} {\partial x \over \partial r} & {\partial y \over \partial r}\\ {\partial x \over \partial \theta} & {\partial y \over \partial \theta}\end{array} \right)=\left( \begin{array}{cc} {\cos(\theta)} & {\sin(\theta)}\\ {-r\sin(\theta)} & {r\cos(\theta)}\end{array} \right)$$

The inverse matrix is

$$\left( \begin{array}{cc} {\partial r \over \partial x} & {\partial \theta \over \partial x}\\ {\partial r \over \partial y} & {\partial \theta \over \partial y}\end{array} \right)=\left( \begin{array}{cc} {\cos(\theta)} & {-\sin(\theta)\over r}\\ {\sin(\theta)} & {cos(\theta)\over r}\end{array} \right)$$

This portion comes from the book "Partial Differential Equations, an Introduction" by Walter Strauss. He also gives a warning that ${\partial r \over \partial x}\not=({\partial x \over \partial r})^{-1}$. I will like to know, under what circumstances can ${dy\over dx}={1 \over {dx \over dy}}$?

Thank you very much for your help.

I will like to know, under what circumstances can $\frac{dy}{dx}=\left(\frac{dx}{dy}\right)^{-1}$?

These are just thoughts around that question: they're too lengthy for comments, but just minor observations.

Suppose you have $u = f(x, y)$, $v = g(x, y)$. Suppose you take point $(x_0, y_0)$ and its' image is $(u_0, v_0)$. Also, let there exist inverse differentiable functions $x = \tilde{f} (u, v)$, $y = \tilde{g}(u, v)$. Ok, you want $\frac{\partial u}{\partial x} = \left ( \frac{\partial x}{\partial u} \right )^{-1}$. We always have an equality for Jacobi matrices: $$\frac{D(u, v)}{D(x, y)}\bigg \vert_{(x_0. y_0)} \frac{D(x, y)}{D(u, v)} \bigg \vert_{(u_0. v_0)} = I $$

What will this imply for a partial derivatives of Jacobi matrices? We'll just check the result of Jacobian mulitplication and compare it with $I$ matrix at right side.
So, the element ${I}_{1, 1} = 1$: since $\frac{\partial u}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial u} = 1$ and $\frac{\partial u}{\partial x} \frac{\partial x}{\partial u} = 1$, you have that $\frac{\partial u}{\partial y}\frac{\partial y}{\partial u} = 0$, which means that $\frac{\partial u}{\partial y} = 0 $ or $\frac{\partial y}{\partial u} = 0$. Let's consider the case $\frac{\partial u}{\partial y} = 0 $; this leads to $\frac{\partial u}{\partial x} \frac{\partial x}{\partial v} = 0$ for ${I}_{1, 2}=0$ and that means $\frac{\partial x}{\partial v} = 0$ (since Jacobi matrices must have full rank). Checking $I_{2,2} = 1$ implies $\frac{\partial v}{\partial y}\frac{\partial y}{\partial v} = 1$. So, you see what necessary conditions appear? They are not very interesting: $\frac{\partial u}{\partial y}(x_0, y_0) = 0 $ or $\frac{\partial y}{\partial u}(u_0, v_0) = 0$ is necessary and sufficient to have $\frac{\partial u}{\partial x} (x_0, y_0) \frac{\partial x}{\partial u} (u_0, v_0) = 1$ and $\frac{\partial v}{\partial y}(x_0, y_0) \frac{\partial y}{\partial v} (u_0, v_0) = 1$.
Something more interesting happens when you want this to be true for all points on plane. This will lead to maps like $u = f(x), \; v = g(x, y)$ (if only one of derivatives $\frac{\partial u}{\partial y}$ or $\frac{\partial y}{\partial u}$ vanishes everywhere) or like $u = f(x), \; v = g(y)$ (if both derivatives vanish everywhere). The latter case is kind of obvious.

It is true if you are dealing with a function of one variable $y = f(x)$.