Solve the following system of equations:

$$
\begin{aligned}
& \left(x^{2}+y^{2}\right) \frac{x}{y}=6 \\
& \left(x^{2}-y^{2}\right) \frac{y}{x}=1
\end{aligned}
$$
First, we can establish that the values $x=0, y=0$ as roots are not admissible. Multiplying (1) by (2), and dividing by $\left[\left(x^{2}-y^{2}\right) \frac{y}{x} \neq 0\right]$, we obtain

$$
x^{4}-y^{4}=6
$$

or

$$
\frac{\left(x^{2}+y^{2}\right) x^{2}}{\left(x^{2}-y^{2}\right) y^{2}}=6
$$

Rearranging the latter equation $(x \neq y)$

$$
x^{4}-5 x^{2} y^{2}+6 y^{4}=0
$$

Equation (4) can also be written as:

$$
\left(x^{2}-2 y^{2}\right)\left(x^{2}-3 y^{2}\right)=0
$$

Thus, either

$$
x^{2}=2 y^{2} \quad \text { and from } \quad(3) \quad y_{\mathrm{I}}^{4}=2, \quad x_{1}^{2}= \pm 2 \sqrt{2}
$$

or

$$
x^{2}=3 y^{2} \quad \text { and from } \quad(3) \quad y_{\mathrm{II}}^{4}=\frac{3}{4}, \quad x_{I I}^{2}= \pm \frac{\sqrt{3}}{2}
$$

Considering the real roots, it is evident from (1) that apart from the first factor on the left side, the $\frac{x}{y}$ factor must also be positive, meaning $x$ and $y$ can only have the same sign.

Thus, the real roots are

From (5)

$$
x_{1}=\sqrt[4]{8}, \quad y_{1}=\sqrt[4]{2} ; \quad x_{2}=-\sqrt[4]{8}, \quad y_{2}=-\sqrt[4]{2}
$$

From (6)

$$
x_{3}=\sqrt[4]{\frac{27}{4}}, \quad y_{3}=\sqrt[4]{\frac{3}{4}} ; \quad x_{4}=-\sqrt[4]{\frac{27}{4}}, \quad y_{4}=-\sqrt[4]{\frac{3}{4}}
$$

Benedek Tomor (Györ, Révai g. III. o. t.)

(Those who are familiar with imaginary numbers can determine from (1) that since the first factor on the left side is negative, $\frac{x}{y}$ must also be negative, meaning the imaginary roots can only pair with opposite signs. Thus, the imaginary roots are:

$$
\begin{aligned}
& x_{5}=i \sqrt[4]{8}, \quad y_{5}=-i \sqrt[4]{2} ; \quad x_{6}=-i \sqrt[4]{8}, \quad y_{6}=i \sqrt[4]{2} \\
& x_{7}=i \sqrt[4]{\frac{27}{4}}, \quad y_{7}=-i \sqrt[4]{\frac{3}{4}} ; \quad x_{8}=i \sqrt[4]{\frac{27}{4}}, \quad y_{8}=i \sqrt[4]{\frac{3}{4}}
\end{aligned}
$$

Substitution can verify that these roots also satisfy our system of equations.)