The second equation factorize as $(y-1)(x-y-1)=0$, subst. $y=x-1$ in the first equation one obtains a quartic which factorizes again as $(3x^2-3x-1)(3x^2-3x+4)=0$ However, here is a different approach (only the "positive" version is shown), which avert the possible difficulties of factorizing a quartic: Let $xy=P, x+y^2=S$, the system becomes $\begin{cases} P^2+P-\frac{4}{9}=0 \\P+1=S\end{cases}\Longrightarrow 9S^2-9S-4=0\Rightarrow S_1=\frac{4}{3}, S_2=-\frac{1}{3}$ $\Longrightarrow x+y^2=\frac{4}{3}, xy=\frac{4}{3}-1=\frac{1}{3}$ $\Longrightarrow\begin{cases} xy=\frac{1}{3} \\x=\frac{4}{3}-y^2\end{cases}\Longrightarrow 3y^3-4y+1=0$ It noted that the cubic is divisible by $y-1\Longrightarrow 3y^3-4y+1=(y-1)(3y^2-3y-1)=0\Longrightarrow y_1=\frac{3+\sqrt{21}}{6}$, $y_2=-\frac{3-\sqrt{21}}{6}$, etc.