7.268. $\left\{\begin{array}{l}9^{\sqrt[4]{x y^{2}}}-27 \cdot 3^{\sqrt{y}}=0, \\ \frac{1}{4} \lg x+\frac{1}{2} \lg y=\lg (4-\sqrt[4]{x}) .\end{array}\right.$

The system of equations is:
\[
\left\{\begin{array}{l}
9^{\sqrt[4]{x y^{2}}}-27 \cdot 3^{\sqrt{y}}=0, \\
\frac{1}{4} \lg x+\frac{1}{2} \lg y=\lg (4-\sqrt[4]{x}) .
\end{array}\right.
\]
Solution.

Domain of definition: $\left\{\begin{array}{l}00 .\end{array}\right.$

Let's write the system of equations as

$\left\{\begin{array}{l}3^{2 \sqrt[4]{x} \sqrt{y}}=3^{3+\sqrt{y}} \\ \lg \sqrt[4]{x}+\lg \sqrt{y}=\lg (4-\sqrt[4]{x})\end{array} \Leftrightarrow\left\{\begin{array}{l}2 \sqrt[4]{x} \cdot \sqrt{y}=3+\sqrt{y}, \\ \lg (\sqrt[4]{x} \cdot \sqrt{y})=\lg (4-\sqrt[4]{x})\end{array} \Leftrightarrow\right.\right.$

$\Leftrightarrow\left\{\begin{array}{l}2 \sqrt[4]{x} \cdot \sqrt{y}=3+\sqrt{y}, \\ \sqrt[4]{x} \cdot \sqrt{y}=4-\sqrt[4]{x}\end{array} \Rightarrow 2=\frac{3+\sqrt{y}}{4-\sqrt[4]{x}} \Leftrightarrow \sqrt{y}=5-2 \sqrt[4]{x}\right.$.

Substituting the value of $\sqrt{y}$ into the first equation, we get $2 \sqrt[4]{x}(5-2 \sqrt[4]{x})=3+5-2 \sqrt[4]{x} \Leftrightarrow(\sqrt[4]{x})^{2}-3 \sqrt[4]{x}+2=0$.

Solving this equation as a quadratic equation in terms of $\sqrt[4]{x}$, we find $\sqrt[4]{x}=1, x_{1}=1$ or $\sqrt[4]{x}=2, x_{2}=16$. Then $y_{1}=9, y_{2}=1$.

Answer: $(1 ; 9),(16 ; 1)$.