Let $z_1$ $z_2$ and $z_3$ non iqual complex numbers and be $\Omega$ an open set such that the 3 points are in the same connected component in $\mathbb{C}-\Omega$, then $$f(z)=\frac{z}{(z-z_1)(z-z_2)(z-z_3)}$$ admits a primitive
We use residue theorem in $\Omega\cup\lbrace z_1,z_2,z_3\rbrace$ and we think the problem is reduced to show that a closed path round the three points the same number of times.
There are complex numbers $\omega_1$, $\omega_2$, and $\omega_3$ such that$$f(z)=\frac{\omega_1}{z-z_1}+\frac{\omega_2}{z-z_2}+\frac{\omega_3}{z-z_3},$$which are$$\omega_1=\frac{z_1}{(z_1-z_2)(z_1-z_3)},\ \omega_2=\frac{z_2}{(z_2-z_1)(z_2-z_3)}\text{, and }\omega_3=\frac{z_3}{(z_3-z_1)(z_3-z_2)}.$$So, if $\gamma\colon[a,b]\longrightarrow\Omega$ is a closed path,$$\oint_\gamma f(z)\,\mathrm dz=\omega_1\operatorname{ind}_\gamma(z_1)+\omega_2\operatorname{ind}_\gamma(z_2)+\omega_3\operatorname{ind}_\gamma(z_3).\tag1$$Since $\Omega$ is connected, the $3$ numbers $\operatorname{ind}_\gamma(z_1)$, $\operatorname{ind}_\gamma(z_2)$, and $\operatorname{ind}_\gamma(z_3)$ are equal; let $j$ be their common value. Then, it follows from $(1)$ that$$\oint_\gamma f(z)\,\mathrm dz=(\omega_1+\omega_2+\omega_3)j=0,$$since $\omega_1+\omega_2+\omega_3=0$. Since this take place for every closed path, $f$ has a primitive.