Solving $z^3+3i\overline{z}=0$ $$z^3+3i\overline{z}=0$$
$z=x+yi$
$$(x+yi)^3+3i(x-yi)=0$$
$$x^3+3x^2yi-3xy^2-y^3i+3ix+3y=0$$
$$x^3-3xy^2+3y=0\text{  and  }  3x^2y-y^3+3x=0$$
How to continue from here?

Multiplying by $z$,
$$z^4+3i|z|^2=0$$ so that $z^4$ is purely imaginary. Then with $\omega$ a fourth root of $-i$,
$$-ir^4+3ir^2=0.$$
We have $r=0\lor r=\sqrt3$ and
$$z=0\lor z=\sqrt 3\,\omega$$ (five solutions in total).
