The graph I want is $$\left\{ \left( {x \over 1 \pm \sqrt{xy}}, {y \over 1 \pm \sqrt{xy}} \right) \in \mathbf R^2 : x^2 + y^2 + xy = 1 \right\}.$$
What I've tried so far is to "rotate by 45 degrees (and scale)," that is, set $x = u+v$ and $y = u-v$. Then the condition $x^2+y^2+xy=1$ becomes $3u^2+v^2=1$, which admits the obvious parametrization $u=\cos(t)/\sqrt3$ and $v=\sin(t)$.
I'm not very satisfied with this parametrization and I was wondering if anyone could come up with a simpler description of what this graph "actually is" (admittedly, all this is entirely subjective).
Not sure if this additional information will help, but secretly, the orange part of the graph is the "inverse" of the blue part of the graph because this is the stereographic projection of two simple closed curves on $S^2$ that are symmetric about the equator. These simple closed curves are the points on $S^2$ that map to the ellipse $x^2+y^2+xy=1$ under the projection $(x,y,z)\mapsto(x,y)$. In other words, the set $\{(x,y,z)\in \mathbf R^3: x^2+y^2+z^2=1,x^2+y^2+xy=1\}$ is a union of two disjoint loops on $S^2$, and the question is: Is there a "simple" description of its image in $\mathbf R^2$ under stereographic projection?
Let me rename your $x$ and $y$ as $s$ and $t$ satisfying $s^2 + t^2 + st = 1$, and let $$ x = \dfrac{s}{1 \pm \sqrt{st}}, \ y = \dfrac{t}{1 \pm \sqrt{st}}$$ With $z = \pm \sqrt{st}$, we have the system of polynomial equations $$ \eqalign{s^2 + t^2 + st - 1 &= 0\cr z^2 - st &= 0\cr x (1+z) - s &= 0\cr y (1+z) - t &= 0\cr}$$ Using Maple, I took a "plex" Groebner basis of the ideal generated by the left sides to eliminate $s, t, z$, and obtained the equation in $x$ and $y$: $$ {x}^{4}+2\,{x}^{2}{y}^{2}+{y}^{4}-2\,{x}^{2}-4\,xy-2\,{y}^{2}+1 = 0$$ This, by the way, is an elliptic curve (genus $1$). Its Weierstrass form is $$ {u}^{3}-{\frac {208\,u}{3}}+{\frac{4480}{27}}+{v}^{2}$$