Insufficient info, maybe?

I took the partial derivatives w.r.t $x$ and $y$ , and set them equal to $0$ . Skipping some steps, this resulted in the equations $x(x+y)^3+(1+xy)(y^2-1)=0$ $y(x+y)^3+(1+xy)(x^2-1)=0.$ Subtracting them results in $(x-y)(x^2+xy+y^2-1)=0$ . If we let $x-y=0$ , then using either condition results in $x^2=\frac{1}{3}$ . Substituting it into the expression (to be minimized), one obtains $2x^2+\frac{(1+x^2)^2}{4x^2}=\frac{2}{3}+\frac{\frac{16}{9}}{\frac{4}{3}}=2.$ If we let $x^2+xy+y^2-1=0$ , then we can substitute $1-xy$ for $x^2+y^2$ anywhere it occurs in the expression (to be minimized), obtaining $x^2+y^2+\frac{(1+xy)^2}{x^2+y^2+2xy}=1-xy+\frac{(1+xy)^2}{1-xy+2xy}=1-xy+\frac{(1+xy)^2}{1+xy}=1-xy+1+xy=2.$ So, therefore, since the expression is strictly positive, with only one discontinuity, an infinite discontinuity, we can see that the global minimum is $2$ .

From the inequality

$(x+y-\frac{1+xy}{x+y})^{2}>0$

one gets

$x^2+y^2+(\frac{1+xy}{x+y})^2>2(x+y)\frac{1+xy}{x+y}-2xy = 2$

By letting $x=1, y=0$ in $x^2+y^2+(\frac{1+xy}{x+y})^2$ One can see that "2" is the solution