Let $S=x+y$ and $P=xy$. $x^2+y^2=2\implies S^2-2P=2$ $\frac{x^2}{2-y}+\frac{y^2}{2-x}=2\implies x^2(2-x)+y^2(2-y)=2(2-x)(2-y)\implies 2(x^2+y^2)-(x^3+y^3)=8-4(x+y)+2xy$ $\implies 2S^2-4P-S^3+3PS=8-4S+2P\implies 2S^2-6P-S^3+3PS+4S-8=0$ $4S^2-6(S^2-2)-2S^3+3(S^2-2)S+8S-16=0\implies S^3-2S^2+2S-4=0\implies(S-2)(S^2+2)=0\implies S\in\left\{2,\pm i\sqrt{2}\right\}$ $P=\frac{S^2}{2}-1\in\{1,-2\}$ $u^2-2u+1=0\implies (x,y)=(1,1)$ $u^2-i\sqrt{2}-2=0\implies (x,y)=\left(\frac{\pm\sqrt{6}+i\sqrt{2}}{2},\frac{\mp\sqrt{6}+i\sqrt{2}}{2}\right)$ $u^2+i\sqrt{2}-2=0\implies (x,y)=\left(\frac{\pm\sqrt{6}-i\sqrt{2}}{2},\frac{\mp\sqrt{6}-i\sqrt{2}}{2}\right)$ The only real solution is $(1,1)$ with multiplicity $2$.