3. [5 points] Solve the system of equations

$$
\left\{\begin{array}{l}
\left(-\frac{x^{7}}{y}\right)^{\ln (-y)}=x^{2 \ln \left(x y^{2}\right)} \\
y^{2}+2 x y-3 x^{2}+12 x+4 y=0
\end{array}\right.
$$
Answer: $(2 ;-2),(3 ;-9),\left(\frac{\sqrt{17}-1}{2} ; \frac{\sqrt{17}-9}{2}\right)$.

Solution. We take the natural logarithm of the first equation in the system:

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

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

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

We rewrite the second equation as $y^{2}+2 y(x+2)-3 x^{2}+12 x=0$ and solve it as a quadratic equation in terms of the variable $y$:

$$
\frac{D}{4}=(x+2)^{2}-\left(12 x-3 x^{2}\right)=x^{2}+4 x+4-12 x+3 x^{2}=(2 x-2)^{2} \Rightarrow y=-(x+2) \pm(2 x-2)
$$

Thus, $y=-3 x$ or $y=x-4$. There are four possible cases:

a) $\left\{\begin{array}{l}y=-x, \\ y=-3 x\end{array} \Leftrightarrow\left\{\begin{array}{l}x=0, \\ y=0 .\end{array} \quad\right.\right.$ The point $(0 ; 0)$ does not satisfy the domain of valid values.

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

c) $\left\{\begin{array}{l}y=-x^{2}, \\ y=-3 x\end{array} \Leftrightarrow\left[\begin{array}{l}\left\{\begin{array}{l}x=0, \\ y=0,\end{array}\right. \\ \left\{\begin{array}{l}x=3, \\ y=-9 .\end{array}\right.\end{array}\right.\right.$ The point $(0 ; 0)$ does not satisfy the domain of valid values.

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

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: $(2 ;-2),(3 ;-9),\left(\frac{\sqrt{17}-1}{2} ; \frac{\sqrt{17}-9}{2}\right)$.