3. [5 points] Solve the system of equations

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

Solution. We take the logarithm of the first equation of the system to base 10:

$$
\lg \left(\frac{x^{4}}{y^{2}}\right) \cdot \lg y=\lg (-x) \cdot \lg (-x y)
$$

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

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

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

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

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

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

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

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

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

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

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

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