Example 2. Solve the system of equations
$$
\left\{\begin{array}{l}
x y+y z+z x=1, \\
y z+z t+t y=1, \\
z t+t x+x z=1, \\
t x+x y+y t=1 .
\end{array}\right.
$$
(3rd Junior High School Mathematics Competition, Jiangsu Province)
Analyzing the original equation, which has many elements and an unclear reduction strategy, we can first explore the system of equations by reducing the number of variables and making an analogy:
$$
\left\{\begin{array}{l}
x y + y z = 1, \\
x y + x z = 1, \\
x z + y z = 1
\end{array}\right.
$$

By subtracting the equations pairwise, we get:
$$
\left\{\begin{array}{l}
z(x - y) = 0, \\
x(y - z) = 0, \\
y(x - z) = 0.
\end{array}\right.
$$
If \( x = 0 \), then \( y z = 0 \), which contradicts \( y z = 1 \). Therefore, \( x \neq 0 \). Similarly, \( y \neq 0 \) and \( z \neq 0 \). Thus, \( x = y = z \). Substituting into the original system of equations, we find \( x = y = z = \pm \frac{\sqrt{2}}{2} \).

By analogy with the above process, we can reduce the original system of equations to the following three equations:
From (1) - (2), we get \( (x - t)(y + z) = 0 \),
From (2) - (3), we get \( (y - x)(z + t) = 0 \),
From (3) - (4), we get \( (z - y)(t + x) = 0 \).
If \( y + z = 0 \), i.e., \( z = -y \). Substituting into (1) gives \( y^2 = -1 \), which has no real solutions, so \( y + z \neq 0 \).
Similarly, \( z + t \neq 0 \) and \( t + x \neq 0 \).
Thus, from (5), (6), and (7), we get:
$$
\begin{array}{l}
x - t = 0, y - x = 0, z - y = 0. \\
\therefore x = y = z = t.
\end{array}
$$

Substituting into (1) gives \( 3x^2 = 1 \),
$$
\therefore x = \pm \frac{\sqrt{3}}{3}.
$$

Thus, the original system of equations has two sets of solutions:
$$
x = y = z = t = \pm \frac{\sqrt{3}}{3}.
$$