Set $u=x+y$ and $v=xy$. We have (a) $z=2-u$ and (b) $v+u(2-u)=1\implies v=(u-1)^2$. Noting $u^2\ge 4v$, we have $u^2\ge (2u-2)^2$, that is $2/3\le u\le 2$. Under this, $|x-y|^2 = u^2-4v = (2-u)(3u-2)$. It is easily seen (e.g. by simple differentiation) that the maxima of parabola $u\mapsto (2-u)(3u-2)$ is at $u=4/3$, which indeed is in $[2/3,2]$. With this, $x-y\le |x-y|\le 2/\sqrt{3}$. The triple finding the equality is easy to find by observing $u=4/3$ and $v=1/9$.

Let $x, y$ be real numbers such that $ xy+(x+y)(2-x-y)=1.$ Prove that $$-\frac 2{\sqrt 3} \leq x-y\leq \frac 2{\sqrt 3}$$$$-\frac 1{8}(5\sqrt 5+1)\leq x-y^2\leq \frac 1{8} (5\sqrt 5-1)$$$$-\frac 2{3} \sqrt{3+2\sqrt 3}\leq x^2-y^2\leq \frac 2{3} \sqrt{3+2\sqrt 3}$$$$-\frac 2{9}\left( 3\sqrt{3+2\sqrt 3}+\sqrt 3\right) \leq x-y^3\leq \frac 2{9}\left( 3\sqrt{3+2\sqrt 3}-\sqrt 3\right)$$

Let $x, y$ be real numbers such that $ xy+(x+y)(3-2x-2y)=1.$ Prove that $$-2\sqrt {\frac{2}{7}}\leq x-y\leq 2\sqrt {\frac{2}{7}}$$$$ \frac{2}{7}(3-\sqrt {2}) \leq x+y\leq \frac{2}{7}(3+\sqrt {2})$$$$-1\leq x^2-y^2\leq 1$$

sqing wrote: Let $x, y$ be real numbers such that $ xy+(x+y)(2-x-y)=1.$ Prove that $$-\frac 2{\sqrt 3} \leq x-y\leq \frac 2{\sqrt 3}$$

Let $x, y$ be real numbers such that $ xy+(x+y)(k+1-kx-ky)=1.$ Prove that $$-2\sqrt{\frac{k^2-2k+2}{4k-1}} \leq x-y\leq 2\sqrt{\frac{k^2-2k+2}{4k-1}}$$Where $k\in N^+.$

Answer:$\frac{2\sqrt3}{3}$ Let's say that $x-y=a,x+y=b$ Then, we have $x=\frac{a+b}{2}$ and $y=\frac{b-a}{2}$ We can see that $x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)=2$ \[(\frac{a+b}{2})^2+(\frac{b-a}{2})^2+z^2=2\]Also we know that $b+z=2\implies z=2-b$ so \[\frac{a^2+b^2}{2}+(2-b)^2=2\]\[a^2+3b^2-8b+8=4\]\[3b^2-8b+(a^2+4)=0\]By discriminant, we get $\Delta=64-12a^2-48=16-12a^2$ So $16\geq 12a^2\implies \frac{2\sqrt3}{3}\geq a$ $x-y=\frac{2\sqrt3}{3},x+y=\frac{4}{3}\implies x=\frac{2+\sqrt3}{3},y=\frac{2-\sqrt3}{3},z=\frac{2}{3}$ holds.