6.114. $\left\{\begin{array}{l}y \sqrt{2 x}-x \sqrt{2 y}=6, \\ x y^{2}-x^{2} y=30\end{array}\right.$
## Solution.

Domain of definition: $\left\{\begin{array}{l}x \geq 0, \\ y \geq 0 .\end{array}\right.$

Rewrite the first equation as $\sqrt{2}\left(\sqrt{x y^{2}}-\sqrt{x^{2} y}\right)=6$ and let $\sqrt{x y^{2}}=u, \sqrt{x^{2} y}=v$, where $u \geq 0, v \geq 0$. Then the system in terms of $u$ and $v$ becomes

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ \sqrt { 2 } ( u - v ) = 6 , } \\
{ u ^ { 2 } - v ^ { 2 } = 3 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ u - v = 3 \sqrt { 2 } , } \\
{ ( u - v ) ( u + v ) = 3 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
u-v=3 \sqrt{2}, \\
u+v=\frac{30}{3 \sqrt{2}}
\end{array} \Leftrightarrow\right.\right.\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
u-v=3 \sqrt{2}, \\
u+v=5 \sqrt{2},
\end{array} \text { from which } u=4 \sqrt{2}, v=\sqrt{2} .\right.
\end{aligned}
$$

Thus, $\left\{\begin{array}{l}\sqrt{x y^{2}}=4 \sqrt{2}, \\ \sqrt{x^{2} y}=\sqrt{2}\end{array} \Leftrightarrow\left\{\begin{array}{l}x y^{2}=32, \\ x^{2} y=2 .\end{array}\right.\right.$

Multiplying these equations, we get $x^{3} y^{3}=64$, from which $x y=4$. Finally, we find $x=\frac{2}{4}=\frac{1}{2}, y=\frac{32}{4}=8$.

Answer: $\left(\frac{1}{2} ; 8\right)$.