Property of a well-known continuous analytic curve is given in polar coordinates as
$$ r(\theta)\cdot r(\theta+\pi)= A $$
where $A$ is constant. Converse of the Proposition 35 of Book 3 of Euclid's Elements, to be now re-established by means of differential equations if not already done before.
How do we find its differential equation in Polar or Cartesian coordinates and solve it?
By product differentiation twice with respect to $\theta$ a second order ODE
$$r(\theta)\cdot r''(\theta + \pi)+r''(\theta )\cdot r(\theta + \pi) +2\; r^{2}(\theta)\cdot r^{'2}(\theta + \pi)=0\quad $$
with two boundary/initial conditions should procure the curve on integration.
Of relevance is may be circle of constant curvature familiar in polar coordinates $r= r(\theta)$ primed on $\theta$:
$$ \dfrac{r^2(\theta ) +2\; r^{'2}(\theta)- r(\theta)\cdot r''(\theta )}{[r^{'2}(\theta)+r^{'2}(\theta)]^\frac32}= \dfrac{ 1}{\sqrt{A^2+\epsilon^2} }$$ where $\epsilon$ is an arbitrary constant.
Appreciate your indication to approach for analytical solution, related links and comments.
You don't. This is not enough information to tell much of anything about the curve.
Let $ f: [0,\pi) \to \Bbb R$ be any function that doesn't take on $0$ as a value. Define $$r(\theta) = \begin{cases} f(\theta),&0\le \theta < \pi\\\dfrac A{f(\theta - \pi)},&\pi\le \theta < 2\pi\end{cases}$$
and extend periodically to all of $\Bbb R$. You can prove that $r$ satisfies the equation.
Since you describe it as a curve, presumably it is continuous, in which case $f$ should be continuous and also satisfy $\lim_{\theta \to \pi-} f(\theta) = A/f(0)$. But this still leaves a vast range of possible $f$. You could even add analytic on as a condition, and still find an uncountably infinite family of functions satisfy your equation.