Let $S=x+y$ and $P=xy$. Taking the sum and product of the two equations and simplifying gives: $\begin{cases} x^2+x+y^2+y=0\\ x^3y^2+x^2y^3+x^3y+x^2y^2+xy^3+x^2y+xy^2+x^2+xy+y^2+x+y=0\\ \end{cases}$ Using Newton's identities gives: $\begin{cases} S^2-2P+S=0\\ P^2S+PS^2-P^2+PS+S^2-P+S=0\\ \end{cases}$ The first equation becomes $P=\frac{S^2+S}{2}$, and substitution gives $4(S^2+S)^2S+2(S^2+S)S^2-(S^2+S)^2+2(S^2+S)S+4S^2-2(S^2+S)+4S=0$. Expand and simplify to get $S^5+3S^4+3S^3+3S^2+2S=0$. Factor to get $S(S+1)(S+2)(S^2+1)=0$. This means $(S,P)\in\left\{(0,0),(-1,0),(-2,1),\left(\pm i,\frac{-1\pm i}{2}\right)\right\}$ The solution to the original system is then the following after discarding extraneous cases. Real: $(0,0)$, $(-1,-1)$ Complex: $(A_+,B_+)$, $(A_-,B_-)$, $(B_+,A_+)$, $(B_-,A_-)$ $A_\pm=-\frac{\sqrt{2+2\sqrt{5}}}{4}\pm\left(\frac{1}{2}+\frac{\sqrt{-2+2\sqrt{5}}}{4}\right)i$ $B_\pm=\frac{\sqrt{2+2\sqrt{5}}}{4}\pm\left(\frac{1}{2}-\frac{\sqrt{-2+2\sqrt{5}}}{4}\right)i$