Solve the following system of equations:

$$
\begin{aligned}
& x y=z(x+y+z) \\
& y z=4 x(x+y+z) \\
& z x=9 y(x+y+z)
\end{aligned}
$$
$$
\begin{aligned}
& x y=z(x+y+z) \\
& y z=4 x(x+y+z) \\
& z x=9 y(x+y+z)
\end{aligned}
$$

Let's denote the value of $x+y+z$ as $t$. If $t=0$, the right-hand side of our equations is 0, so one of the unknowns must be 0. It cannot be that only one of the unknowns is 0, because then the left-hand side of the equation in which this unknown does not appear could not be 0. However, if two of the unknowns are 0, then the value of the third must also be 0, since their sum is 0. Therefore, if $t=0$, the only possibility is

$$
x=y=z=0
$$

which is indeed a solution to the system of equations. We will show that if $t \neq 0$, then none of the unknowns can be 0. For example, if $x$ were 0, then according to (2), one of the other two would also be 0, but then the left-hand sides would all be 0, which can only happen if all three variables are 0, which is not possible because $t \neq 0$.

By multiplying the equations in pairs, we get

$$
\begin{aligned}
& x y^{2} z=4 x z t^{2} \\
& x y z^{2}=36 x y t^{2} \\
& x^{2} y z=9 y z t^{2} .
\end{aligned}
$$

Since we only need to determine the roots when $t \neq 0$, the values of the unknowns cannot be 0, so we can simplify by them. Taking the square root, we get

$$
y= \pm 2 t, z= \pm 6 t, x= \pm 3 t .
$$

If $z=6 t$, then $x+y+z \geq-3 t-2 t+6 t=t$, and the equality sign can only hold if $x=-3 t, y=-2 t$. If $z=-6 t$, then $x+y+z \leq 3 t+2 t-6 t=-t$, so the sum cannot be $t$. Therefore, we get that only

$$
x=-3 t, y=-2 t, z=6 t
$$

can be considered. By substitution, we can verify that these values satisfy our system of equations for any $t$. Since (5) also gives the root of (4) when $t=0$, (5) provides all the solutions to the system of equations we are examining.