9. (16 points) Given real numbers $x, y$ satisfy $2^{x}+2^{y}=4^{x}+4^{y}$, try to find the range of values for $U=8^{x}+8^{y}$.
9. Let $a=2^{x}, b=2^{y}$, then the given condition can be transformed into $a+b=a^{2}+b^{2}$. Completing the square, we get
$$
\left(a-\frac{1}{2}\right)^{2}+\left(b-\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2},
$$

According to equation (1) in the $a O b$ plane, it is easy to see that $t=a+b \in(1,2]$.
Also, $a b=\frac{(a+b)^{2}-\left(a^{2}+b^{2}\right)}{2}=\frac{t^{2}-t}{2}$, thus
$$
U=8^{x}+8^{y}=a^{3}+b^{3}=(a+b)^{3}-3 a b(a+b)=t^{3}-3 \cdot \frac{t^{2}-t}{2} \cdot t=-\frac{1}{2} t^{3}+\frac{3}{2} t^{2} .
$$

Let $f(t)=-\frac{1}{2} t^{3}+\frac{3}{2} t^{2}$, then when $t \in(1,2]$,
$$
f^{\prime}(t)=-\frac{3}{2} t^{2}+3 t=-\frac{3}{2} t(t-2)>0 .
$$

Thus, $f(t)$ is monotonically increasing on $(1,2]$. It is easy to see that when $t \in(1,2]$, $f(t) \in(1,2]$.
In conclusion, the range of $U=8^{x}+8^{y}$ is $(1,2]$.