I'll paraphrase what my ODE textbook says (sadly I don't have a pdf).
Essentially, suppose that a differential equation is given in the form of $y'=f(x,y)$, where $f$ is defined on $\mathcal{D}\subset \mathbb{R}^2$. We can now define a reciprocal ODE as $\displaystyle x'=\frac{1}{f(x,y)}$. From here we can define the notion of domain of an ODE. (Definition is quoted from the textbook)
Definition Domain of an ordinary differential equation $y'=f(x,y)$ (that is, its reciprocal ordinary differential equation) is the union of domain of functions $f(x,y)$ and $\frac{1}{f(x,y)}$.
You should also keep in mind that domain doesn't contain points at which the functions $f(x,y)$ and $\frac{1}{f(x,y)}$ are undefined.
Here is an example.
If $\displaystyle y'=\frac{y\ln{y}}{\sin{x}}$, then $\displaystyle x'=\frac{\sin{x}}{y\ln{y}}$. The domain $\mathcal{D}_1$ of the function $\displaystyle f(x,y)$ is $\mathcal{D}_1=\{(x,y)\in \mathbb{R}^2:\sin{x}\neq 0,y\in(0,+\infty)\}=(\cup_{k\in\mathbb{Z}}(k\pi,(k+1)\pi))\times(0,+\infty)$ and the domain $\mathcal{D}_2$, of the function $\displaystyle \frac{1}{f(x,y)}$ is $\displaystyle \mathcal{D}_2=\{(x,y)\in \mathbb{R}^2:x\in\mathbb{R},y\in(0,1)\cup(1,+\infty)\}=(-\infty,+\infty)\times((0,1)\cup(1,+\infty))$.
Clearly the functions aren't defined at points $(k\pi,1), k\in\mathbb{Z}$, hence these points should be taken out of the domai. So we get the domain $\mathcal{D}=\mathcal{D}_1\cup\mathcal{D}_2=((-\infty,+\infty)\times(0,+\infty))\backslash\{(k\pi,1):k\in\mathbb{Z}\}$.
Since this is pretty much the first things that is mentioned in the textbook I can't continue anything without understanding this. So my questions are the following two:
$1.$ Why is $x'$ defined as $\frac{1}{f(x,y)}$? Is it even defined or can we define it?
$2.$ What is the intuition behind domain being defined in such a way?
I couldn't find anything online about either of these things.
It seems that this approach tries to incorporate vertical solutions or solutions with vertical points from the start. Usually this is covered in a ver general way in the discussion of so-called exact differential equations $$M(x,y)\,dx-N(x,y)\,dy=0.$$
Practically that means that you can separate the terms in $f(x,y)$ as components of the fraction $\frac{M(x,y)}{N(x,y)}$, if possible in a variant without common roots. Then the curve in the $(x,y)$ plane is also obtained as the trajectory of the system $$\dot x(t)=N(x(t),y(t)),\\\dot y(t)=M(x(t),y(t)).$$
In the extreme cases one can obtain $t=x$ where $(M,N)=(f,1)$ and $t=y$ where $(M,N)=(1,1/f)$.
In my opinion this is not a good way to start. It allows to include some special situations, but makes the exposition harder. And in the switch to systems there just is no reciprocal of the vector field. My approach in identifying a common denominator, or in finding a reparametrization that avoids being singular at singular points, still works, but should be left as extension of some basic theory.