Hello, Albert!
I have a very looong solution.
Find solutions of $x,y,z$
.\begin{array}{cccc}\frac {1}{x}+\frac {1}{y+z} &=& \frac {1}{2} & [1[ \\<br />
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\frac {1}{y}+\frac {1}{z+x}&=&\frac {1}{3} & [2] \\<br />
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\frac {1}{z}+\frac {1}{x+y}&=&\frac {1}{4} & (3) \end{array}
Note that: .x,y,z\, \ne\,0.
\begin{array}{cccccccccc}[1]\!:\;2y + 2x + 2x \:=\:x(y+z) &\Rightarrow& x + y + z \:=\: \frac{xy+xz}{2} & [4] \\<br />
[2]\!:\; 3x + 3x + 3y \:=\:y(x+z) &\Rightarrow & x+y+z \:=\:\frac{xy+yz}{3} & [5] \\<br />
[3]\!:\; 4x+3y+4z \:=\:z(x+y) & \Rightarrow & x+y+z \:=\:\frac{xz+yz}{4} & [6] \end{array}
\text{From }[4],[5],[6]\!:\;\underbrace{\frac{xy+xz}{2}}_{[7]} \:=\:\underbrace{\frac{xy + yz}{3}}_{[8]} \:=\:\underbrace{\frac{xz+yz}{4}}_{[9]}
[7]=[8]\!:\;3xy + 3xz \:=\:2xy + 2yz \;\;\;\Rightarrow\;\;\; x \:=\:\frac{2yz}{y+3x}\;\;[10]
[8] = [9]\!:\;4xy + 4yz \:=\:3xz + 3yz \;\;\;\Rightarrow\;\;\;x \:=\:\frac{yz}{3x-4y}\;\;[11]
[10]=[11]\!:\;\frac{2yz}{y+3x} \:=\:\frac{yz}{3z-4y} \quad\Rightarrow\quad z \:=\:3y\;\;[12]
Substitute into [10]: .x \:=\:\frac{2y(3y)}{y+3(3y)} \quad\Rightarrow\quad x \:=\:\tfrac{3}{5}y\;\;[13]
Substitute [12] and [13] into [4]:
.. 2y + 2(3y) + 2\left(\tfrac{3}{5}y\right) \;=\; \left(\tfrac{3}{5}y\right)y + \left(\tfrac{3}{5}y\right)(3y)
. . \tfrac{46}{5}y \:=\:\tfrac{12}{5}y^2 \quad\Rightarrow\quad 6y^2\:=\:23y \quad\Rightarrow\quad \boxed{y \:=\:\tfrac{23}{6}}
Substitute onto [13}: .x \:=\:\tfrac{3}{5}\left(\tfrac{23}{6}\right) \quad\Rightarrow\quad \boxed{x \:=\:\frac{23}{10}}
Substitute into [12]: .z \:=\:3\left(\tfrac{23}{6}\right) \quad\Rightarrow\quad \boxed{z \:=\:\frac{23}{2}}