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 $n>3$
$$x^{2^n}+y^{2^n}>\frac 2{3^{2^n}} \text{?}$$

1) to show $x^4+y^4 < 8$, you already have a good solution by squaring $x^2+y^2<2+xy$. 
2) By AM-GM and Cauchy Schwarz, we get
$$\left(x^4+y^4+\frac{x^4+y^4}2\right)(3) \ge (x^4+y^4+x^2y^2)(1^2+1^2+(-1)^2) \ge (x^2+y^2-xy)^2 > 1$$
hence $x^4+y^4 > \frac29$
For the more general case, you could use Power Means in addition to the above, i.e. for $n>2$:
$$\sqrt[2^n]{\frac{x^{2^n}+y^{2^n}}2} \ge \sqrt[4]{\frac{x^4+y^4}2} > \frac1{\sqrt3} \implies x^{2^n}+y^{2^n} > \frac2{3^{2^{n-1}}}$$
which is a tighter bound than your inequality.
