How can I prove that $2/9$x$ and $y$ are real numbers.
Given that $1<x^2-xy+y^2<2$, how can I show that $\frac 29<x^4+y^4<8$ ?
Then can I use that to prove that for any natural number $n>3$
$$x^{2^n}+y^{2^n}>\frac 2{3^{2^n}} \text{?}$$

$x^2 - xy + y^2 = 1$ is an ellipse. to find out the minor and major axis, we change variables $$x = \xi \cos t - \eta \sin t, y = \xi \sin t + \eta \cos t$$ where $t$ will be determined later.
the ellipse in $\xi, \eta$ coordinates is $$\xi^2(\cos^2 t -\sin t \cos t+\sin^2 t) + \xi \eta(-2\cos t \sin t+\sin^2 t - \cos^2 t+2\sin t \cos t) + \eta^2(\sin ^2 t +\sin t \cos t +\cos ^2 t) =1 $$
set $t = \frac{\pi}{4},$ you get $$\frac12\xi ^2+\frac32\eta^2= 1. $$
so the major and minor axis of the ellipse $x^2 - xy + y^2 = 1$ are 
$\sqrt 2$ and $ \sqrt { \frac23}.$ 
therefore by symmetry the minimum of$x^4 + y^4$ on the hyperbola $x^2 -xy + y^2 = 1$ is achieved on the minor axis $x = -\sqrt\frac13, y = \sqrt \frac 13$ and the minimum value is $\dfrac{2}{9}.$
therefore by symmetry the minimum of$x^4 + y^4$ on the hyperbola $x^2 -xy + y^2 = 2$ is achieved on the major axis $x = \sqrt 2, y = \sqrt 2$ and the maximum value is $8.$
we have established $$ \dfrac{2}{9} \le x^4 + y^4 \le 8 \text { on } 1 \le x^2 - xy + y^2 \le 2.$$
