\begin{align*} 2A'+3A^2\bar{A}&=0\\ 2\bar{A}'+3\bar{A}^2A&=0 \end{align*} with ICs $:A(0)=-\frac{\textit{i}}{2}, \bar{A}(0)=\frac{i}{2}$
The solution is given \begin{align*}A(t)=-\frac{i}{\sqrt{4+3t}}\\\bar{A}(t)=\frac{i}{\sqrt{4+3t}} \end{align*}
How was this solution calculated?
One has $$ \frac{d}{dt}|A|^2=2Re(\bar AA')=3Re(\bar A^2A^2)=3|A|^4 $$ Thus $$ |A|^2=\frac{1}{|A(0)|^{-2}+3t}=\frac1{4+3t} $$ Now insert this into the first equation $$ 2A'=3|A|^2A=\frac{3}{4+3t}A $$ and solve as linear or separable equation.