We can eliminate $y$ and factor the resulting equation to get $x \left(x^2-x-1\right) \left(x^6-2 x^5+2 x^4-2 x^3+2 x^2-x+1\right)$ The linear and quadratic factor are due to the fact that, real or complex, the system must include any solution with $x=y$. However, the factor $x^6-2 x^5+2 x^4-2 x^3+2 x^2-x+1=0$ gives no real solutions since you can write it as \[(x^3-x^2)^2+(x^2-x)^2+\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=0.\] So the real solutions are all with$x=y$ and are $(0,0)$, $\left(\frac{1\pm\sqrt{5}}{2},\frac{1\pm\sqrt{5}}{2}\right)$ Note that it is very possible, by just letting $S=x+y$ and $P=xy$, to solve for the complex solutions in closed form, although doing that takes a lot of work. The complex solutions are: $(A_\pm,A_\mp)$, $(B_\pm,C_\pm)$, $(C_\pm,B_\pm)$ where \begin{align*} A_\pm&=\frac{1}{3}-\frac{1}{12} \left(\sqrt[3]{44+12 \sqrt{69}}-\sqrt[3]{-44+12 \sqrt{69}}\right)\pm\frac{i}{12} \sqrt{-24+6 \sqrt[3]{1484-132 \sqrt{69}}+6 \sqrt[3]{1484+132 \sqrt{69}}}\\ \\ B_\pm&=\frac{1}{3}+\frac{1}{24} \left(\sqrt[3]{44+12 \sqrt{69}}-\sqrt[3]{-44+12 \sqrt{69}}+\sqrt{48+6 \sqrt[3]{1484-132 \sqrt{69}}+6 \sqrt[3]{1484+132 \sqrt{69}}+12 \sqrt{-84+6 \sqrt[3]{26668-2508 \sqrt{69}}+6 \sqrt[3]{26668+2508 \sqrt{69}}}}\right)\\ &\pm\frac{i}{24} \left(\sqrt{3} \left(\sqrt[3]{44+12 \sqrt{69}}+\sqrt[3]{-44+12 \sqrt{69}}\right)-\sqrt{-48-6 \sqrt[3]{1484-132 \sqrt{69}}-6 \sqrt[3]{1484+132 \sqrt{69}}+12 \sqrt{-84+6 \sqrt[3]{26668-2508 \sqrt{69}}+6 \sqrt[3]{26668+2508 \sqrt{69}}}}\right)\\ \\ C_\pm&=\frac{1}{3}+\frac{1}{24} \left(\sqrt[3]{44+12 \sqrt{69}}-\sqrt[3]{-44+12 \sqrt{69}}-\sqrt{48+6 \sqrt[3]{1484-132 \sqrt{69}}+6 \sqrt[3]{1484+132 \sqrt{69}}+12 \sqrt{-84+6 \sqrt[3]{26668-2508 \sqrt{69}}+6 \sqrt[3]{26668+2508 \sqrt{69}}}}\right)\\ &\pm\frac{i}{24} \left(\sqrt{3} \left(\sqrt[3]{44+12 \sqrt{69}}+\sqrt[3]{-44+12 \sqrt{69}}\right)+\sqrt{-48-6 \sqrt[3]{1484-132 \sqrt{69}}-6 \sqrt[3]{1484+132 \sqrt{69}}+12 \sqrt{-84+6 \sqrt[3]{26668-2508 \sqrt{69}}+6 \sqrt[3]{26668+2508 \sqrt{69}}}}\right) \end{align*}

@rchokler kudos to you for that last section.

$0=x^3-x^2-y-(y^3-y^2-x)=(x-y)(x^2+xy+y^2-(x+y)+1)=(x-y)\frac{(x-1)^2+(y-1)^2+(x+y)^2}{2} \to x=y \to x^3-x^2-x=0 \to x=y=0,x=y=\frac{1 \pm \sqrt{5}}{2}$