3. Solve the system of equations
$$
\left\{\begin{array}{l}
\frac{9-4^{y+1}-3^{x} \times 2^{y+2}-9^{x}}{4^{y+1}+3^{x} \times 2^{y+1}-3^{x+1}-9}=\frac{3^{x}+2^{y+1}-3}{3^{x}-1} \\
3^{x+1} \times 2^{y+1}=1
\end{array}\right.
$$
3. Let the variable substitution be $a=3^{x}, b=2^{y+1}(a>0, b>0)$.

Then, the second equation transforms to $a b=3$.
The first equation becomes
$$
\begin{array}{l}
\frac{9-a^{2}-2 a b-b^{2}}{b^{2}+a b-3 a-9}=\frac{a+b-3}{a-1} \\
\Rightarrow \frac{(3-a-b)(3+a+b)}{(b-3)(a+b+3)}=\frac{a+b-3}{a-1} \\
\Rightarrow \frac{3-a-b}{b-3}=\frac{a+b-3}{a-1} .
\end{array}
$$

We will discuss two cases below.
(1) If $a+b=3$, then from $a b=3$, we get
$$
a^{2}-3 a+3=0,
$$

this equation has no solution;
(2) If $b-3=1-a$, then from $a b=3$, we solve to get $(a, b)=(3,1)$ or $(1,3)$ (the latter is discarded).
Thus, $x=1, y=-1$.
In summary, $x=1, y=-1$.