Solution. From the definition of the logarithmic function, it follows that $y-x^{3}>0$, $y-x^{3} \neq 1$, $x^{3}+y>0$, and $x^{3}+y \neq 1$. Furthermore, if we use that $\log _{b} a=\frac{1}{\log _{a} b}$, the given system is equivalent to the system:

$$
\left\{\begin{array}{l}
2^{y-x^{3}} \log _{y+x^{3}}\left(y-x^{3}\right)=1 \\
2^{y-x^{3}} 3^{y-x^{3}} \log _{x^{3}+y}\left(y-x^{3}\right)=9
\end{array}\right.
$$

If we substitute from the first equation into the second, we sequentially obtain:

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ 2 ^ { y - x ^ { 3 } } \operatorname { l o g } _ { y + x ^ { 3 } } ( y - x ^ { 3 } ) = 1 , } \\
{ 2 ^ { y - x ^ { 3 } } 3 ^ { y - x ^ { 3 } } \operatorname { l o g } _ { x ^ { 3 } + y } ( y - x ^ { 3 } ) = 9 , }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
2^{y-x^{3}} \log _{y+x^{3}}\left(y-x^{3}\right)=1, \\
3^{y-x^{3}}=3^{2},
\end{array}\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ 2 ^ { y - x ^ { 3 } } \operatorname { l o g } _ { y + x ^ { 3 } } ( y - x ^ { 3 } ) = 1 , } \\
{ y - x ^ { 3 } = 2 , }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ 2 ^ { 2 } \operatorname { l o g } _ { y + x ^ { 3 } } 2 = 1 , } \\
{ y - x ^ { 3 } = 2 , }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\log _{y+x^{3}} 2^{4}=1, \\
y-x^{3}=2,
\end{array}\right.\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ y + x ^ { 3 } = 1 6 , } \\
{ y - x ^ { 3 } = 2 , }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ 2 y = 1 8 , } \\
{ 2 x ^ { 3 } = 1 4 , }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ y = 9 , } \\
{ x ^ { 3 } = 7 , }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
y=9, \\
x=\sqrt[3]{7} .
\end{array}\right.\right.\right.\right.
\end{aligned}
$$

Finally, since for the found values of $x$ and $y$ the conditions

$$
y-x^{3}>0, y-x^{3} \neq 1, x^{3}+y>0 \text { and } x^{3}+y \neq 1
$$

are satisfied, the system has a unique solution $x=\sqrt[3]{7}, y=9$.
1. In the set of real numbers, solve the system of equations

$$
\left\{\begin{array}{l}
\log _{y-x^{3}}\left(x^{3}+y\right)=2^{y-x^{3}} \\
\frac{1}{9} \log _{x^{3}+y}\left(y-x^{3}\right)=6^{x^{3}-y}
\end{array}\right.
$$