7.263. $\left\{\begin{array}{l}4^{\frac{x}{y}+\frac{y}{x}}=32, \\ \log _{3}(x-y)=1-\log _{3}(x+y)\end{array}\right.$

The system of equations is:
\[
\left\{\begin{array}{l}
4^{\frac{x}{y}+\frac{y}{x}}=32, \\
\log _{3}(x-y)=1-\log _{3}(x+y)
\end{array}\right.
\]
## Solution.

Domain of definition: $\left\{\begin{array}{l}x \neq 0, \\ y \neq 0, \\ x-y>0, \\ x+y>0 .\end{array}\right.$

Rewrite the system of equations as

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ 2 ^ { \frac { 2 x } { y } + \frac { 2 y } { x } } = 2 ^ { 5 } , } \\
{ \operatorname { log } _ { 3 } ( x - y ) = \operatorname { log } _ { 3 } \frac { 3 } { x + y } }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ \frac { 2 x } { y } + \frac { 2 y } { x } = 5 , } \\
{ x - y = \frac { 3 } { x + y } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
2\left(\frac{x}{y}\right)+\frac{2}{\left(\frac{x}{y}\right)}-5=0, \\
x^{2}-y^{2}=3
\end{array} \Leftrightarrow\right.\right.\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
2\left(\frac{x}{y}\right)^{2}-5\left(\frac{x}{y}\right)+2=0, \\
x^{2}-y^{2}=3
\end{array} \Rightarrow \frac{x}{y}=\frac{1}{2} \text { or } \frac{x}{y}=2 .\right.
\end{aligned}
$$

If $y=2 x$, then $x^{2}-4 x^{2}=-3 x^{2} \neq 3$. Let $x=2 y$, then the given system is equivalent to two systems:

1) $\left\{\begin{array}{l}y=\frac{x}{2}, \\ x=-2\end{array} \Leftrightarrow\left\{\begin{array}{l}x=-2, \\ y=-1\end{array}\right.\right.$ do not satisfy the domain of definition;
2) $\left\{\begin{array}{l}y=\frac{x}{2}, \\ x=2\end{array} \Leftrightarrow\left\{\begin{array}{l}x=2, \\ y=1\end{array}\right.\right.$

Answer: $(2 ; 1)$.