Problem 6. We will "flip" each of the equations in the system:

$$
\left\{\begin{array}{l}
\frac{x+y}{x y}=1 \\
\frac{y+z}{y z}=\frac{1}{2} \\
\frac{x+z}{x z}=\frac{1}{3}
\end{array}\right.
$$

(The transformation is equivalent since none of the right-hand sides can become zero).

Notice that $\frac{x+y}{x y}=\frac{1}{x}+\frac{1}{y}$, and so on. Therefore, we have obtained a system of linear equations in terms of $\frac{1}{x}, \frac{1}{y}$, and $\frac{1}{z}$.

Solving it, we get $\frac{1}{x}=\frac{5}{12}, \frac{1}{y}=\frac{7}{12}, \frac{1}{z}=-\frac{1}{12}$.
Answer. $x=\frac{12}{5}, y=\frac{12}{7}, z=-12$