Differentiating with respect to arc length $s$ ( upto sign of derivative) the relation ( here it is explicit but sufficient for further work).
Consider right triangle with $PA$ as hypotenuse.
$$ \boxed{\sin \omega = \frac{y}{d}} ,\; \quad\cos \omega \frac{ d\omega}{ds}= \frac{dy/ds}{d} \tag
0 $$
By definition of curvature and the right triangle,
$$ \frac{\cos \omega }{PC1=R_1}= \frac{\sin \omega}{d} ;\; R_1 = d \cot \omega \tag 1 $$
where $R_1=PC1 $ is the principal/major radius of curvature of the Tractrix and serves as the equator corresponding .
Next, the length of normal vector from P upto x-axis.
$$ \frac{y}{\cos \omega} = \frac{d \sin \omega}{\cos \omega} = R_2 \text{ where } R_2=PC2 = d \tan \omega \tag 2 $$
$R_2$ is the second principal minor radius of curvature in a direction perpendicular the tangent at $P.$
The product of radii of curvatures obtained by multiplying relations (1) and (2):
$$ R_1\cdot R_2 = d^2 \tag 3 $$
which is a more common ode form for the Tractrix.
At the cusp point $P_0,\;$ $R_1=0, R_2= \infty$
When the rotation ally formed surface is considered
$$ k_1= -1/R_1,\; k_2= 1/R_2,\; K= -k_1 k_2=-1/d^2.\; \tag 4 $$
which is the constant negative Gaussian curvature of the Pseudosphere surface.
Since the curvature occur on opposite sides of the tractrix meridian $R_1$ is considered negative by convention with respect to a common normal.
Alternately (0) can be differentiated,
$$ \cos \omega \;k_1 = \frac{\sin \omega}{d};\; k_1 k_2 = \frac {\tan \omega }{d} \frac{\cos \omega }{y}= \frac {\tan \omega }{d} \frac{\cos \omega }{\sin \omega/d}$$
$$ K= \frac{1}{d^2}$$
The boxed relation is mentioned in several text books of Differential geometry, including Blashke, Eisenhardt. I recommend including it in the cited German Wiki now if possible.
EDIT1:
Feel I understand your question better now.
How do the set of circles on x-axis result in the differential equation of the tractrix? If this is the question then the required condition/clue is supplied in the same diagram you have included.
Radius PA is the tangent length and is always perpendicular to the instantaneous position of the circle arc with radius PA at P. In other words what is the orthogonal trajectory of these sliding circles traced by P? is what I thought was the question in other words.
The procedure is standard. You can take it from here. Find its ode, eliminate moving parameter $a$ between them, replace $ y' \to \;1/y' {\text or } \;\omega \to \;\pi/2 - \omega $ for orthogonal trajectory and then integrate. Fix one of the integrated solutions at cusp points as initial condition.
This also results in (4) the characteristic ode of the Tractrix of hyperbolic geometry.
Best Answer
Your approach is valid, but it's easy to lose ourselves in the notation. I will make the substitution $g = \frac{dy}{dx}$.
Equation $(A)$ becomes $4x^3 + 4y^3 g = 0$. Differentiating with respect to $y$, we have
$$12x^3 + 12y^2\frac{dy}{dx}g+4y^3\frac{dg}{dx}=0.$$
Notice that another $\frac{dy}{dx}$ popped up from $\frac{d}{dx}4y^3$. Now, $$12x^3 + 12y^2\left(\frac{dy}{dx}\right)^2+4y^3\frac{d^2y}{dx^2}=0.$$
Solving this for $\frac{d^2y}{dx^2}$ will yield the correct result after some manipulation.