4. [ $[$ points] Solve the system of equations

$$
\left\{\begin{array}{l}
\frac{6}{x^{2}+y^{2}}+x^{2} y^{2}=10 \\
x^{4}+y^{4}+7 x^{2} y^{2}=81
\end{array}\right.
$$
Answer: $(\sqrt{3} ; \pm \sqrt{3}),(-\sqrt{3} ; \pm \sqrt{3})$.

Solution. Introduce new variables $u=x^{2}+y^{2}, v=x^{2} y^{2}$. Then $x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2}=u^{2}-2 v$ and the system takes the form

$$
\left\{\begin{array}{l}
\frac{6}{u}+v=10 \\
u^{2}+5 v=81
\end{array}\right.
$$

From the first equation $v=10-\frac{6}{u}$; substitute this expression into the second equation and solve it:

$$
\begin{aligned}
& u^{2}+5\left(10-\frac{6}{u}\right)=81 \Leftrightarrow u^{2}-31-\frac{30}{u}=0 \Leftrightarrow \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ u ^ { 3 } - 3 1 u - 3 0 = 0 , } \\
{ u \neq 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ ( u + 1 ) ( u + 5 ) ( u - 6 ) = 0 , } \\
{ u \neq 0 }
\end{array} \Leftrightarrow \left[\begin{array}{l}
u=-1 \\
u=-5 \\
u=6
\end{array}\right.\right.\right.
\end{aligned}
$$

Since $u>0$, only the value $u=6$ is suitable. Therefore, $v=9$, and in the original variables we get

$$
\left\{\begin{array} { l } 
{ x ^ { 2 } + y ^ { 2 } = 6 , } \\
{ x ^ { 2 } y ^ { 2 } = 9 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x^{2}=6-y^{2} \\
\left(6-y^{2}\right) y^{2}=9
\end{array} \Leftrightarrow x^{2}=y^{2}=3\right.\right.
$$

From this, it follows that the system has 4 solutions: $(\sqrt{3} ; \pm \sqrt{3}),(-\sqrt{3} ; \pm \sqrt{3})$.