3. $[5$ points] Solve the system of equations

$$
\left\{\begin{array}{l}
\left(x^{2} y^{4}\right)^{-\ln x}=y^{\ln \left(y / x^{7}\right)} \\
y^{2}-x y-2 x^{2}+8 x-4 y=0
\end{array}\right.
$$
Answer: $(2 ; 2),(2 ; 4),\left(\frac{\sqrt{17}-1}{2} ; \frac{9-\sqrt{17}}{2}\right)$.

Solution. We take the logarithm of the first equation of the system with base $e$:

$$
\ln \left(x^{2} y^{4}\right) \cdot(-\ln x)=\ln y \cdot \ln \left(y / x^{7}\right)
$$

This equation on the domain of valid values is equivalent to the following:

$$
\begin{aligned}
& (2 \ln x+4 \ln y) \ln x=\ln y(7 \ln x-\ln y) \Leftrightarrow \ln ^{2} y-3 \ln x \ln y+2 \ln ^{2} x=0 \Leftrightarrow \\
& (\ln y-\ln x)(\ln y-2 \ln x)=0 \Leftrightarrow(y-x)\left(y-x^{2}\right)=0 \Leftrightarrow\left[\begin{array}{l}
y=x \\
y=x^{2}
\end{array}\right.
\end{aligned}
$$

We rewrite the second equation as $y^{2}-y(x+4)-2 x^{2}+8 x=0$ and solve it as a quadratic equation in terms of the variable $y$:

$$
D=(x+4)^{2}-4\left(8 x-2 x^{2}\right)=x^{2}+8 x+16-32 x+8 x^{2}=(3 x-4)^{2} \Rightarrow y=\frac{(x+4) \pm(3 x-4)}{2}
$$

Thus, $y=2 x$ or $y=4-x$. There are four possible cases:

a) $\left\{\begin{array}{l}y=x, \\ y=2 x\end{array} \Leftrightarrow\left\{\begin{array}{l}x=0, \\ y=0 .\end{array}\right.\right.$ The point $(0 ; 0)$ does not satisfy the domain of valid values.

b) $\left\{\begin{array}{l}y=x, \\ y=4-x\end{array} \Leftrightarrow\left\{\begin{array}{l}x=2 \\ y=2 .\end{array}\right.\right.$

c) $\left\{\begin{array}{l}y=x^{2}, \\ y=2 x\end{array} \Leftrightarrow\left[\begin{array}{l}\left\{\begin{array}{l}x=0 \\ y=0\end{array}\right. \\ \left\{\begin{array}{l}x=2, \\ y=4 .\end{array}\right.\end{array}\right.\right.$

The point $(0 ; 0)$ does not satisfy the domain of valid values.

d) $\left\{\begin{array}{l}y=x^{2}, \\ y=4-x\end{array} \Leftrightarrow\left\{\begin{array}{l}y=4-x, \\ x^{2}+x-4=0\end{array} \Leftrightarrow\left\{\begin{array}{l}y=4-x, \\ x=\frac{-1 \pm \sqrt{17}}{2}\end{array} \Leftrightarrow\left[\begin{array}{l}\left\{\begin{array}{l}x=\frac{-1+\sqrt{17}}{2}, \\ y=\frac{9-\sqrt{17}}{2},\end{array}\right. \\ \left\{\begin{array}{l}x=\frac{-1-\sqrt{17}}{2}, \\ y=\frac{9+\sqrt{17}}{2} .\end{array}\right.\end{array}\right.\right.\right.\right.$

The point $\left(\frac{-1-\sqrt{17}}{2} ; \frac{9+\sqrt{17}}{2}\right)$ does not satisfy the domain of valid values.

Combining the results, we get the final answer: $(2 ; 2),(2 ; 4),\left(\frac{\sqrt{17}-1}{2} ; \frac{9-\sqrt{17}}{2}\right)$.