Answer: $(1, 1, 1), (-1, -1, -1)$.

Numbers $x$, $y$ and $z$ must have the same sign, because if exactly one or two of them are negative then there exists an equation in the system whose all terms in the l.h.s. are negative and cannot sum up to $3$. It is also easy to see that $(x, y, z)$ being a solution implies $(-x, -y, -z)$ being a solution, too.
Hence, w.l.o.g., assume that $x$, $y$ and $z$ are positive. Subtracting the second equation from the first one, the third equation from the second one, and the first equation from the third one, we obtain the following new system:
$$\begin{cases} x \left(y + \frac{1}{y}\right) = z \left(y + \frac{1}{x}\right), \\ y \left(z + \frac{1}{z}\right) = x \left(z + \frac{1}{y}\right), \\ z \left(x + \frac{1}{x}\right) = y \left(x + \frac{1}{z}\right). \end{cases}$$
The first equation of the new system shows that $x > z$ holds if and only if $y + \frac{1}{y} < y + \frac{1}{x}$, where the latter inequality obviously holds if and only if $y > x$. Similarly, the second equation implies that $y > x$ if and only if $z > y$, and the third equation implies that $z > y$ if and only if $x > z$. Thus any of the inequalities $x > z$, $y > x$ and $z > y$ yields the impossible cycle $x > z > y > x$. Consequently, we must have $x \le z \le y \le x$ which implies $x = y = z$. Every equation of the original system now reduces to $1+1+x^2 = 3$. Hence $x = y = z = 1$.
Besides the positive solution, the system has the corresponding negative solution $(-1, -1, -1)$.