Clearly $x,y$, and $z$ are all non-zero: otherwise at least two of the fractions would be $0$. Thus, we may invert everything and rewrite the system as
$$\left\{\begin{align*} &\frac{x+y}{xy}=1\\\\ &\frac{x+z}{xz}=\frac12\\\\ &\frac{y+z}{yz}=\frac13\;. \end{align*}\right.$$
Now divide out on the lefthand side:
$$\left\{\begin{align*} &\frac1x+\frac1y=1\\\\ &\frac1x+\frac1z=\frac12\\\\ &\frac1y+\frac1z=\frac13\;. \end{align*}\right.$$
Now substitute $u=\dfrac1x,v=\dfrac1y$, and $w=\dfrac1z$ to get
$$\left\{\begin{align*} &u+v=1\\\\ &u+w=\frac12\\\\ &v+w=\frac13\;. \end{align*}\right.$$
This is a very easy system to solve, and once you’ve solved it, you can easily get $x,y$, and $z$.