This ODE has some very interesting properties. If one clears fractions and writes it out as
$$
x(x+2y)(x-2y+1)\,y'' = (4x^2-8y^2+3x+4y)\,y' + x(4y-1)\,(y')^2,
\tag1
$$
one recognizes this as the equation for geodesics of a projective connection in the complement $D$ (which has $7$ components) of the three lines $x = 0$, $x+2y=0$, and $x-2y+1=0$ in the $xy$-plane. Moreover, because the right hand side of (1) has no terms of degree $0$ or $3$ in $y'$, it follows that the lines $x=x_0$ and $y=y_0$ are geodesics of this projective connection in $D$ and should be regarded as 'solutions' of the equation. This is probably most evident if one writes the equation in parametric form for a curve $\bigl(x(t),y(t)\bigr)$, in which case, the equation becomes
$$
x(x{+}2y)(x{-}2y{+}1)\,\bigl(\dot y\,\ddot x-\dot x\,\ddot y\bigr) + x(4y{-}1)\,\dot x\,\dot y^2 + (4x^2{-}8y^2{+}3x{+}4y)\,\dot x^2\,\dot y =0
\tag2
$$

Contrary to what the OP claims, there is a $2$-parameter family of solutions that are regular at $x=0$. If one looks for a formal power series solution in the form
$$
y(x) = a_0 + a_1\,x + a_2\,x^2 + a_3\, x^3 + \cdots,\tag3
$$

then one finds, by examining the three lowest terms in the equation, that one must have either
$$
(i)\ \ a_1 = a_2 = 0,\qquad (ii)\ \ a_0 = 0,\ a_1 = -1,
\quad\text{or}\quad (iii)\ \ a_0=\tfrac12,\ a_1 = 1.
$$
Since $y(x)$ is a solution if and only if $\tfrac12 - y(x)$ is a solution, the second and third cases are essentially the same, so I will treat only the first two cases henceforth.

In the first case, one finds that there is a formal power series solution of the form
$$
y(x) = \tfrac14(1{+}a) + \frac{(a^2{-}1)b}{12} x^3 -\frac{b}{4}\,x^4 - \frac{b}{5}\,x^5 - \frac{a(a^2{-}1)b^2}{72}\,x^6 + \cdots + p_k(a,b)\,x^k + \cdots,\tag4
$$
where $p_k(a,b)$ is a (unique) polynomial in constants $a$ and $b$. Moreover, this series has a positive radius of convergence for each $(a,b)$. [Proofs that these and similar series listed below have positive radii of convergence can be based on techniques in the book *Singular Nonlinear Partial Differential Equations* by R. Gérard and H. Tahara.] Note that the symmetry $y(x)\mapsto \tfrac12 - y(x)$ corresponds to the symmetry $(a,b)\mapsto (-a,-b)$.

In the second case (and, similarly, via the symmetry $y\mapsto \tfrac12 - y$, the third case), one finds that there is a formal power series solution
$$
y(x) = - x + \frac{b}{2}\,x^2 - \frac{b}{5}\,x^3 + \frac{b(b{+}3)}{10}\,x^4
- \frac{4b(13b{+}25)}{175}\,x^5 + \cdots + q_k(b)\,x^k + \cdots,
\tag5
$$
where $q_k(b)$ is a (unique) polynomial in $b$ of degree at most $\tfrac12 k$. This series has a positive radius of convergence for every $b$. The value $b=0$ gives the solution $y(x) = -x$ and the value $b=-5/4$ gives the solution $y(x) = \tfrac14 - \tfrac14(1+3x)(1+2x)^{1/2}$. Note that (4) with $a=-1$ and (5) give two distinct $1$-parameter families of solutions passing through the point $(x,y)=(0,0)$, where the two singular lines $x=0$ and $x+2y=0$ meet.

As for analytic solutions meeting the singular line $x+2y=0$, there exist two distinct $2$-parameter families of series solutions: The first is given in parametric form by
a formal power series
$$
\begin{aligned}
x(t) &= a + a(2a{+}1)\,t\,,\\
y(t) &= -\frac{a}{2} + a(2a{+}1)b\,t^2\left(1 + \frac{2(5a{-}4b{+}2)}{3}\,t +\cdots + p_k(a,b)\,t^k + \cdots \right),
\end{aligned}
\tag6
$$
where $p_k(a,b)$ is a (unique) polynomial in $a$ and $b$ and where the $y$-series in $t$ has a positive radius of convergence for every $(a,b)$. The second $2$-parameter family can be written in the form
$$
\begin{aligned}
x(t) &= a + a(2a{+}1)\,t^3\,,\\
y(t) &= -\frac{a}{2} + (2a{+}1)\,t^2\left(b + a\,t + \frac{2b^2}{5}\,t^2 +\cdots + q_k(a,b)\,t^k + \cdots \right),
\end{aligned}
\tag7
$$
where $q_k(a,b)$ is a (unique) polynomial in $a$ and $b$ and where the $y$-series in $t$ has a positive radius of convergence for every $(a,b)$. Interestingly, these solutions with $b\not=0$ have a cusp singularity at $t=0$, while, when $b=0$, only the terms involving $t^{3k}$ remain, so that $x(t)$ and $y(t)$ are analytic functions of $t^3$.

Note, however, that these two series solutions degenerate at the special values $a=0$ and $a=-\tfrac12$. The value $a = 0$ corresponds to the point $(x,y)=(0,0)$, where the singular lines $x=0$ and $x+2y=0$ cross, while the value $a=-\tfrac12$ corresponds to the point $(x,y) = (-\tfrac12,\tfrac14)$, where the singular lines $x+2y=0$ and $x-2y+1=0$ cross.

Finally, through the singular point $(x,y) = (-\tfrac12,\tfrac14)$, where the singular lines $x+2y=0$ and $x-2y+1=0$ cross, there are two convergent series solutions with one parameter: The first is
$$
\begin{aligned}
x(t) &= -\frac{1}{2} + t\,,\\
y(t) &= +\frac{1}{4} + b\,t^3 -3b\,t^4 + \cdots + f_k(b)\,t^k + \cdots ,
\end{aligned}
\tag8
$$
where $f_k(b) = -f_k(-b)$ is a polynomial in $b$. The second series is
$$
\begin{aligned}
x(t) &= -\frac{1}{2} + b\,t^2 + \frac{b^2(5b{+}32)}{16}\,t^4 + \cdots + g_k(b)\,t^{2k} + \cdots\,,\\
y(t) &= +\frac{1}{4} + t\, ,
\end{aligned}
\tag9
$$
where $g_k(b)$ is a polynomial in $b$. Note that the value $b=2$ in this latter series corresponds to the known solution(s) represented by solving $(y-\tfrac14)^2-(1+3x)^2(1+2x) = 0$ for $y$ as a function of $x$.

It is very interesting that through every point of the $xy$-plane, there passes at least one $1$-parameter family of solution curves, and, along two of the singular lines, there can be two distinct $1$-parameter families of solution curves.

One more interesting feature should be noted: Because the equation defines a projective structure on $D$, each solution curve in $D$ comes equipped with a canonical projective structure, i.e., a 'developing map' to $\mathbb{RP}^1$ that is unique up to linear fractional transformation and provides a local parametrization of the curve. This developing map extends analytically across the points where such a curve crosses one of the three singular lines, but the developing map is no longer a local diffeomorphism at such places; its differential vanishes to second or third order at such places.

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