11. (20 points) Given real numbers $x, y$ satisfy $3^{x}+3^{y}=9^{x}+9^{y}$.
Find the range of $U=27^{x}+27^{y}$.
11. Let $a=3^{x}, b=3^{y}$. Then the given equation can be transformed into
$$
\begin{array}{l}
a+b=a^{2}+b^{2}(a, b>0) \\
\Rightarrow\left(a-\frac{1}{2}\right)^{2}+\left(b-\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}
\end{array}
$$

From the graph of equation (1) in the $a O b$ plane, we know that
$$
t=a+b \in(1,2] \text {. }
$$

Also, $a b=\frac{(a+b)^{2}-\left(a^{2}+b^{2}\right)}{2}=\frac{t^{2}-t}{2}$, then
$$
\begin{array}{l}
U=27^{x}+27^{y}=a^{3}+b^{3} \\
=(a+b)^{3}-3 a b(a+b) \\
=t^{3}-3 \times \frac{t^{2}-t}{2} \cdot t=-\frac{1}{2} t^{3}+\frac{3}{2} t^{2} .
\end{array}
$$

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

Thus, $f(t)$ is monotonically increasing on $t \in(1,2]$. It is easy to see that when $t \in(1,2]$, $f(t) \in(1,2]$. In conclusion, the range of $U=27^{x}+27^{y}$ is $(1,2]$.