Does $\frac{d x}{dy}$ indicate that $x=f(y)$? Does the expression $\dfrac{d x}{dy}$ indicate that $x=f(y)$ ? Can we compute it even if $y=g(x)$ ?
For example, if $y=g(x)=x^2,$ is $\dfrac{d x}{dy}$ meaningful even though $y=g(x),$ rather than $x=f(y)$ ?
 A: $y=g(x)$ establishes a functional relation between $x$ and $y$, and under some conditions, it can be inverted as $x=g^{-1}(y)$. Hence there is no real difference between $y=g(x)$ and $x=f(y)$.
This said, when the derivatives exist, the following relation holds:
$$\frac{dx}{dy}=\frac1{\dfrac{dy}{dx}}.$$

Illustration:
For positive $x$ and $y$,
$$y=x^2\iff x=\sqrt y.$$
Then using the common rules of differentiation,
$$\frac{dy}{dx}=2x$$
and
$$\frac{dx}{dy}=\frac1{2\sqrt y}=\frac1{2x}.$$
A: In a case like this, I would hope most authors would provide a little bit of context (e.g. "let $x:\mathbb R \to \mathbb R$ be a function such that..." etc.) as it would be very confusing otherwise! However given no additional information, we can still deduce a few things.
The derivative as Leibniz used it was defined in terms of infinitesimals. In this sense, $dy$ and $dx$ were well defined quantities, and $\frac{dy}{dx}$ was actually a fraction, not just notation. By this definition, which is nowadays a nonstandard approach to calculus (though an equally intriguing one), $$\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}},$$ as our fraction rules would suggest.
In modern mathematics, infinitesimals have largely been abandoned in the field of analysis (calculus' big brother), and therefore $\frac{d}{dx}$ is really just an alternative notation for the derivative function. Formally, the derivative function takes a differentiable function as an input, and gives its derivative as the output. If we let $D$ be our derivative function, and $f$ be a differentiable function, $f' := D(f)$. By this definition, derivatives only make sense in terms of functions. If $y=f(x)$, we use a little bit of slight of hand, and define $$\frac{dy}{dx} = D(f).$$ Therefore, if we suppose that $y = f(x)$, $\frac{dx}{dy}$ has no defined meaning. Since $x$ is not defined as a function of $y$, we do not have a function to plug into our derivative function $D$, and therefore the notation is meaningless.
It can potentially make sense however when $f^{-1}$ exists, in which case, $x = f^{-1}(y)$, and $\frac{dx}{dy} = D(f^{-1})$. This is not super explicit so I don't like it very much, but yes it does work. Try as an exercise to see what this derivative will equal.
Therefore, by our modern definition, $\frac{dx}{dy}$ only makes sense when $x=f(y)$.
P.S. To Theo Diamantakis (I don't have enough reputation to comment), the case of $x=f(z)$ could be defined as you say. Your definition seems to be taking the partial derivative of some implicit function $g(y, z) = f(z)$ with respect to $y$. This seems sensible at first, but it's actually rather stupid to do. It's imprecise and confusing. Instead of doing the above, I would hope that whenever something like this arises in context, the author would be explicit about what all the notations mean, and not randomly take the derivative of $x$ with respect to something other than which it was "defined". It was bad enough that this author swapped $y$ and $x$ in the first place. You can debate all day about what this notation should mean, but the reality is that something like that should not happen.
A: In Essence , $dy/dx$ tells us how $y$ changes when $x$ changes , while $dx/dy$ tell us how $x$ changes when $y$ changes.
In general , these are reciprocals.
"Does $dx/dy$ imply x=f(y) .... not the other way around" : No , It will not imply $x=f(y)$ nor will it imply $y=g(x)$ , though these are the common cases in high school introductory texts. It DOES imply that the 2 variables are inter-related , not which variable is Independent & which variable is Dependant.
More-over , there are not 2 Possibilities , there is a third Possibility , which is more general :
$h(x,y)=0$
Even in that case , we can calculate $dy/dx$ & $dx/dy$ , which will neither imply $y=f(x)$ nor imply $x=g(y)$. It DOES tell us that the variables are inter-related , with $h(x,y)=0$ in most general terms. In certain cases , it might get simplified.
Case 3 $h(x,y)=0$ :
Consider $h(X,Y) = Y + \sin Y - X \cos X = 0$
Plot given by wolfram :

We can calculate (by using rules of Differentiation) that :
$dy/dx + \cos Y dy/dx - \cos X + X \sin X = 0$
which gives us
$dy/dx = (\cos X - X \sin X)/(1 + \cos Y) \tag{1}$
$dx/dy = (1 + \cos Y)/(\cos X - X \sin X) \tag{2}$
Here (1) & (2) are Entirely Equivalent , though it will neither imply $y=f(x)$ nor imply $x=g(y)$ , where we know that $h(x,y)=0$
Case 1 $y=f(x)$ (which is Equivalent to Case 2 $x = y(x)$ ) :
Consider easier case :
$y = e^{+9x}$
$dy/dx = 9e^{+9x} \tag{3}$
$dx/dy = e^{-9x}/9 \tag{4}$
It can be changed to
$x = (\log y)/9$
$dx/dx = 1/9y \tag{5}$
$dy/dx = 9y \tag{6}$
Here (3) , (4) , (5) , (6) are all Entirely Equivalent.
We have calculated both $dy/dx$ & $dx/dy$ in 2 variations each , though we can not know which is Independent variable & which is Dependent variable.
Summary : In all these Cases , we can & we are able to calculate both $dy/dx$ & $dx/dy$ . . . . .
It tell us how one variable changes when the other variable changes , nothing less , nothing more.
