Is $ \frac{\mathrm{d}{x}}{\mathrm{d}{y}} = \frac{1}{\left( \frac{\mathrm{d}{y}}{\mathrm{d}{x}} \right)} $? In calculus, is $ \dfrac{\mathrm{d}{x}}{\mathrm{d}{y}} = \dfrac{1}{\left( \dfrac{\mathrm{d}{y}}{\mathrm{d}{x}} \right)} $? I’m so confused about this matter. What would be a proof of it?
Edit: By the Chain Rule, $ \dfrac{\mathrm{d}{y}}{\mathrm{d}{x}} \cdot \dfrac{\mathrm{d}{x}}{\mathrm{d}{y}} = \dfrac{\mathrm{d}{y}}{\mathrm{d}{y}} = 1 $, so this confuses me.
 A: If the real variables $x$ and $y$ are dependent functionally on each other, say by the equations $y = f(x)$ and $x = g(y)$, then where things are differentiable, we do indeed have an equation
$$ \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} $$
and the most straightforward proof is as you indicated by the chain rule:
$$ \frac{dy}{dx} \frac{dx}{dy} = \frac{dy}{dy} = 1$$
and then solving the equation
$$ \frac{dy}{dx} \frac{dx}{dy} = 1$$
for $\frac{dy}{dx}$.
The argument can be rephrased in terms of functions rather than dependent variables: from the identity
$$ x = g(f(x)) $$
we differentiate to obtain
$$ 1 = g'(f(x)) f'(x) $$
and thus the analogous identity
$$ f'(x) = \frac{1}{g'(f(x))} $$

More generally, in the language of differential forms, the expression $\frac{dy}{dx}$ is defined to be the expression satisfying
$$ dy = \frac{dy}{dx} dx $$
if that is possible. When both $\frac{dy}{dx}$ and $\frac{dx}{dy}$ are defined, then we have two equations
$$ dy = \frac{dy}{dx} dx \qquad \qquad dx = \frac{dx}{dy} dy $$
and can substitute to obtain
$$ dy = \frac{dy}{dx} \frac{dx}{dy} dy $$
and so we again have
$$\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} $$
wherever $dy$ is nonvanishing. An example of a situation where both are defined is if the variables $x$ and $y$ are functionally related by a differentiable equation $f(x,y) = 0$. Taking the differential gives
$$ f_1(x,y) dx + f_2(x,y) dy = 0$$
where $f_1$ is the derivative of the two-variable function $f$ with respect to the first variable. When both $f_1$ and $f_2$ are nonzero, we can solve the equation in both ways to see that both $\frac{dx}{dy}$ and $\frac{dy}{dx}$ are defined.
A: The precise statement is as follows.

Let $ I $ be an open interval, and suppose that $ f: I \to \mathbb{R} $ is one-to-one and continuous on $ I $. If $ f $ is differentiable at $ a \in I $ and $ f'(a) \neq 0 $, then $ f^{-1}: f[I] \to I $ is differentiable at $ b = f(a) $ and
  $$
(f^{-1})'(b) = \frac{1}{f'(a)}.
$$

Using Leibniz’s notation, the formula above can be expressed as
$$
\frac{dx}{dy} \Bigg|_{y = b} = \frac{1}{\left( \dfrac{dy}{dx} \Bigg|_{x = a} \right)}.
$$
