3. In the field of real numbers, solve the system of equations

$$
x^{2}=\frac{1}{y}+\frac{1}{z}, \quad y^{2}=\frac{1}{z}+\frac{1}{x}, \quad z^{2}=\frac{1}{x}+\frac{1}{y} .
$$
SOLUTION. If the numbers $x, y, z$ are a solution to the given system, it is clear that $x y z \neq 0$. Therefore, multiply each equation by the factors $y z, z x$ and $x y$ respectively, and solve the equivalent system of equations in the domain of non-zero real numbers:

$$
x^{2} y z=y+z, \quad x y^{2} z=x+z, \quad x y z^{2}=x+y.
$$

By adding the left and right sides of this system of equations, we obtain the equation after rearrangement:

$$
(x y z-2)(x+y+z)=0.
$$

From this, we see that $x y z=2$ or $x+y+z=0$.

- Let $x y z=2$. By substituting the product $x y z$ in the system (1), we get

$$
2 x=y+z, \quad 2 y=x+z, \quad 2 z=x+y
$$

which is equivalent to the system

$$
3 x=x+y+z, \quad 3 y=x+y+z, \quad 3 z=x+y+z.
$$

From this, it follows that $x=y=z$. Considering the condition $x y z=2$, we get $x=y=z=\sqrt[3]{2}$. By verification, we confirm that the triplet $(\sqrt[3]{2}, \sqrt[3]{2}, \sqrt[3]{2})$ is indeed a solution to the system (1), and thus to the original system of equations.

- Let $x+y+z=0$. From the first equation of the system (1), we get $x^{2} y z=-x$, and with the condition $x \neq 0$, we obtain $x y z=-1$. Verify that every triplet of non-zero real numbers $(x, y, z)$ satisfying the system of two equations

$$
x+y+z=0, \quad x y z=-1
$$

is a solution to the original system. From the equalities (2), it follows that

$$
\frac{1}{y}+\frac{1}{z}=\frac{y+z}{y z}=\frac{-x}{-1 / x}=x^{2}
$$

(considering the symmetry of the given system, it was sufficient to verify one equation).

The system of equations (2) has infinitely many solutions in the domain of non-zero real numbers, which we can obtain, for example, by choosing one variable (e.g., $z$) as a parameter. This gives us the system

$$
x+y=-z, \quad x y=-\frac{1}{z}.
$$

By substituting $x$ from the first equation into the second, we get

$$
(y+z) y=\frac{1}{z}
$$

i.e.,

$$
y^{2}+y z-\frac{1}{z}=0.
$$

This is a quadratic equation in $y$ with parameter $z$. Its discriminant is $D=z^{2}+4 / z$. A necessary and sufficient condition for this equation to have real roots is the inequality $D \geq 0$. Solving the inequality $\left(z^{3}+4\right) / z \geq 0$, we get the condition for the parameter $z$:

$$
z \in(-\infty,-\sqrt[3]{4}\rangle \cup(0, \infty)
$$

Under condition (4), the quadratic equation (3) has roots

$$
y_{1}=\frac{-z+\sqrt{z^{2}+4 / z}}{2} \quad \text{and} \quad y_{2}=\frac{-z-\sqrt{z^{2}+4 / z}}{2}
$$

to which, according to the relation $x=-y-z$, correspond the values

$$
x_{1}=\frac{-z-\sqrt{z^{2}+4 / z}}{2} \quad \text{and} \quad x_{2}=\frac{-z+\sqrt{z^{2}+4 / z}}{2}
$$

with only in the case $z=-\sqrt[3]{4}$ it holds that $\left(x_{1}, y_{1}\right)=\left(x_{2}, y_{2}\right)$.

Conclusion. The given system has the solution $x=y=z=\sqrt[3]{2}$. All other solutions are triplets $(x, y, z)$ of the form

$$
(x, y, z)=\left(\frac{-z \pm \sqrt{z^{2}+4 / z}}{2}, \frac{-z \mp \sqrt{z^{2}+4 / z}}{2}, z\right)
$$

where $z$ is any number satisfying condition (4).

GUIDING PROBLEM:

Solve the system of equations in the domain of real numbers:

$$
x=\frac{1}{y}+\frac{1}{z}, \quad y=\frac{1}{z}+\frac{1}{x}, \quad z=\frac{1}{x}+\frac{1}{y}.
$$

$[x=y=z=\sqrt{2}$. The equalities $x=y=z$ follow from the fact that all three sums $x+y, x+z, y+z$ are equal to each other; their value is $x y z$.]