= Solution
Write the <rational function> $A$ in reduced form. We first exclude a finite pole in its dependent variable at a generic point $(a,b)$. If this pole has order $m\ge1$, the inverse equation
$$
\frac{dz}{dw}=\frac1{A(z,w)},\qquad z(b)=a,
$$
has a <holomorphic> right-hand side near $(a,b)$, with a zero of order $m$ in $w-b$ at $z=a$. Choosing a generic point avoids collisions of pole branches and zeros of the reduced numerator. The unique inverse solution has
$$
z(w)-a=c(w-b)^{m+1}+O((w-b)^{m+2}),\qquad c\ne0.
$$
To justify the leading order, first note $z'(b)=0$; if the first nonzero term has order $\ell$, substitution in the inverse equation makes the right-hand side start at order $m$, since the change caused by $z-a$ has order at least $\ell$. Thus $\ell-1=m$. Inverting produces an <algebraic branch point> of order $m+1$. Away from the fixed coefficient exceptions, $a$ can be chosen throughout an open set on the pole curve; inverse solutions based at these different $a$ give different initial values at a nearby ordinary point. These are <movable singularities of a complex differential equation>, contradicting the hypothesis. Hence the denominator has no genuine $w$-dependent factor: $A$ is a <polynomial> in $w$, with <rational functions> of $z$ as coefficients.
Let its degree in $w$ be $d$. Use the dependent coordinate $v=1/w$ on the <Riemann sphere>. Its equation is
$$
v'=-v^2A(z,1/v).
$$
If $d\ge3$, the right-hand side has a pole of order $d-2$ at $v=0$ for generic $z$. The same inverse-equation argument gives a movable <algebraic branch point> of order $d-1$. It follows that $d\le2$, so \b[the equation is Riccati], including its linear and constant special cases:
$$
\boxed{f'=a(z)f^2+b(z)f+c(z).}
$$
The requirement about the whole dependent-variable <Riemann sphere> is essential: examining only finite values would miss the branching at $f=\infty$ for degree at least three.
The <linearization of a Riccati equation> is particularly symmetric in the system
$$
\binom{f_1}{f_2}'=
\begin{pmatrix}b/2&c\\-a&-b/2\end{pmatrix}
\binom{f_1}{f_2}.
$$
On a simply connected domain where the coefficients are <holomorphic>, its solutions are <holomorphic>. Direct differentiation gives
$$
\left(\frac{f_1}{f_2}\right)'=
\frac{(bf_1/2+cf_2)f_2-f_1(-af_1-bf_2/2)}{f_2^2}
=a\left(\frac{f_1}{f_2}\right)^2+b\frac{f_1}{f_2}+c.
$$
A nonzero initial vector never becomes the zero vector, by uniqueness for the <linear differential equation>, so the quotient is <meromorphic>. Choosing initial vector $(f(z_0),1)$ represents every finite initial value, and $(1,0)$ represents infinity. At fixed coefficient singularities, a single globally <holomorphic> vector solution need not exist; the assertion is local on ordinary domains, or on their appropriate continuation surface.
Finally, for any two finite solutions $u,v$ of the <Riccati equation>,
$$
(u-v)'=(a(u+v)+b)(u-v).
$$
Use this identity on the four differences in
$$
K=\frac{(f-g)(h-j)}{(f-j)(h-g)}.
$$
The <logarithmic derivative> of this <cross-ratio> is
$$
a(f+g+h+j-f-j-h-g)+b+b-b-b=0.
$$
Thus $K$ is constant wherever the expression is initially defined, and the identity extends as an identity of <meromorphic functions>. Distinctness of $g,h,j$ is preserved at ordinary points by uniqueness, interpreted on the <Riemann sphere>. Solving the constant <cross-ratio> equation gives the <Riccati cross-ratio superposition>
$$
\boxed{f=\frac{K(h-g)j-(h-j)g}{K(h-g)-(h-j)}.}
$$
For $K=\infty$ this means $f=j$; $K=0$ gives $g$, and $K=1$ gives $h$. The constant is selected by the initial value of $f$, so the formula represents every solution, with its denominator zeros interpreted as <poles>.
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