7.77 $\left\{\begin{array}{l}\left(x^{2}+y\right) 2^{y-x^{2}}=1, \\ 9\left(x^{2}+y\right)=6^{x^{2}-y} .\end{array}\right.$

The system of equations is:
\[
\left\{\begin{array}{l}
\left(x^{2}+y\right) 2^{y-x^{2}}=1, \\
9\left(x^{2}+y\right)=6^{x^{2}-y} .
\end{array}\right.
\]
7.77 Multiplying both sides of the second equation by $6^{y-x^{2}}>0$, we get the system

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

Dividing the second equation by the first term by term:

$9 \cdot 3^{y-x^{2}}=1$, or $3^{y-x^{2}+2}=3^{0}$,

from which $y-x^{2}+2=0$, i.e., $y-x^{2}=-2$. It remains to solve the system of equations

$$
\left\{\begin{array} { l } 
{ y - x ^ { 2 } = - 2 , } \\
{ ( x ^ { 2 } + y ) \cdot 2 ^ { - 2 } = 1 , }
\end{array} \text { or } \left\{\begin{array}{l}
y-x^{2}=-2 \\
y+x^{2}=4
\end{array}\right.\right.
$$

We solve this system algebraically by addition: $2 y=2$,
i.e., $y=1 ; 2 x^{2}=6$, i.e., $x_{1,2}= \pm \sqrt{3}$.

Answer: $(\sqrt{3} ; 1)(-\sqrt{3} ; 1)$.