The following is the formal statement of a conjecture that feels almost obvious, but I cannot find a reference for it. The idea is that one can obtain the integral curves of a vector field $V(z)$ by locally following curves with the same initial direction. Below, a dot indicate derivative wrt the (real) parametrization variable.
Conjecture:
Let $V(z)$ be a rational vector field. For every $u \in \mathbb{C}$, suppose have the following:
if $u$ is not a pole of $V$, and $V(u)\neq 0$, there is an analytic curve $\gamma_u:[0,T) \to \mathbb{C}$ starting at $u$, with the property that $\dot{\gamma_u}(0) = V(u)$.
if $u$ is a pole of $V$ of order $m$, then there are $m+1$ analytic curves $\gamma_{u,j}:[0,T) \to \mathbb{C}$, $j=1,2,\dotsc,m+1$ starting at $u$, such that \begin{equation} (\dot{\gamma}_{u,j}(0))^{m+1} = \lim_{z \to u} V(z)(z-u)^m. \tag{$\ast$} \end{equation} Moreover, all the values $\dot{\gamma}_{u,j}(0)$ have different arguments which pairwise differ by an multiple of $2\pi/(m+1)$.
Now, let $M$ be any closed set with the property that \begin{equation} u \in M \implies \begin{cases} \gamma_u \subseteq M &\text{ if $u$ is not a pole of $V$},\\ \gamma_{u,1},\dotsc,\gamma_{u,m+1}\subseteq M &\text{ if $u$ is a pole of order $m$ of $V$}. \end{cases} \end{equation} Then,
if $u \in M$ is not a pole or zero of $V$, then $M$ contains the integral trajectory of $V$ starting at $u$.
if $u \in M$ is a pole of $V$, then $M$ contains all the separatrices of $V$ originating at $u$.
A separatrix is an integral curve $\gamma$ of $V$ starting at a pole of $V$ and satisfying ($\ast$).
Idea: Away from poles, then this should be easy, by using Euler's step method. However, the technical issue is the behavior near the poles. Note that the only integral curves actually passing through poles of $V$ are the separatrices.
