= Solution
The <Painlevé property> says that the general solution has no movable critical singularities: <analytic continuation> around a singularity whose location depends on the initial data must not create a different branch. A <fixed singularity of a complex differential equation> has its location prescribed by the coefficients; a <movable singularity of a complex differential equation> has a location which varies with the initial data. A commonly used stronger form requires movable singularities to be <poles>, so that the solution is locally a <meromorphic function> there. Movable <branch points> are excluded; a movable <pole> is permitted. We will prove the stronger form for this equation by finding its complete general solution family as <rational functions>.
Use the <linearization of a Riccati equation>. If $w'=aw^2+bw+c$, put $w=-v'/(av)$ wherever $a,v\ne0$. Differentiating this expression and cancelling the quadratic terms yields
$$
v''-\left(\frac{a'}a+b\right)v'+acv=0.
$$
Conversely a nonzero solution $v$ of this <linear ordinary differential equation> gives a solution $w$, wherever the quotient is defined. Every local finite solution $w$ at an <ordinary point> arises this way: solve $v'=-awv$ with a nonzero initial value and differentiate to get the second-order equation. Multiplying $v$ by a nonzero constant leaves $w$ unchanged, giving the one effective <constant of integration> expected for a first-order <Riccati equation>.
For the present coefficients, $a'/a=1/z+1/(z+1)$, so the <linear ordinary differential equation> simplifies to
$$
z(z+1)v''+(z-1)v'-v=0.
$$
Two direct substitutions give solutions $v_1=z-1$ and $v_2=1/(z+1)$. Their <Wronskian> is
$$
v_1v_2'-v_1'v_2=-\frac{2z}{(z+1)^2},
$$
which is nonzero at every <ordinary point> $z\notin\{0,-1\}$. They therefore form a fundamental pair there, and all local solutions have the form
$$
v=C_1(z-1)+\frac{C_2}{z+1}
=\frac{C_1z^2+C_2-C_1}{z+1},\qquad (C_1,C_2)\ne(0,0).
$$
Taking its <logarithmic derivative> gives
$$
w=\frac{1}{z(z+1)^2}-\frac{2C_1}{(z+1)(C_1z^2+C_2-C_1)}.
$$
Thus, for $C_1\ne0$, writing $k=(C_2-C_1)/C_1$, the general family is
$$
\boxed{w(z)=\frac{1}{z(z+1)^2}-\frac{2}{(z+1)(z^2+k)},\qquad k\in\mathbb C.}
$$
The remaining member, corresponding to $C_1=0$, is
$$
\boxed{w(z)=\frac{1}{z(z+1)^2}.}
$$
The completeness of the linearizing pair proves that no other local solution is omitted. These are <rational functions>, single-valued <meromorphic functions>; hence their continuation has no <branch points>. Their only possible moving singularities are zeros of $z^2+k$. A root $z_0$ outside $\{0,-1\}$ is simple, and its <residue> is
$$
\operatorname*{Res}_{z=z_0}w=-\frac{1}{z_0(z_0+1)}.
$$
Equivalently, <poles of a Riccati solution from zeros of its linearizing solution> are simple at <ordinary points>, because a nontrivial solution of a <linear ordinary differential equation> cannot have both $v(z_0)=0$ and $v'(z_0)=0$. The exceptional collisions $k=0$ or $k=-1$ occur at the fixed coefficient singularities, and still give <rational functions>. Consequently every movable singularity is a <pole>, establishing the <Painlevé property> in its stronger meromorphic sense.
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