Lost in X, Y, Z

Using the usual lowercase, and let $\color{#3D99F6}{u = 2y}$ and $\color{#D61F06}{v = 4z}$ :

$\begin{cases} x + \color{#3D99F6}{2y} + \color{#D61F06}{4z} = 9 & & \Rightarrow x + \color{#3D99F6}{u} + \color{#D61F06}{v} = 9 \\ 4yz + 2zx + xy = 13 & \Rightarrow (2y)(4z)+4zx + x(2y) = 2(13) & \Rightarrow uv + xu + vx = 26 \\ xyz = 3 & \Rightarrow x(2y)(4z) = 8(3) & \Rightarrow uvx = 24 \end{cases}$

Using Vieta's Formulas , this implies that $u$ , $v$ and $x$ are roots of:

$\begin{aligned} w^3 - 9w^2+26w-24 & = 0 \\ (w-2)(w-3)(w-4) & = 0 \end{aligned}$

$\Rightarrow (x,u,v) = \begin{cases} (2, 3,4) & \Rightarrow (x,y,z) = \left(2,\frac{3}{2},1 \right) & \color{#3D99F6}{\text{acceptable solution}} \\ (2,4,3) & \Rightarrow (x,y,z) = \left(2,2,\frac{3}{4} \right) & \color{#3D99F6}{\text{acceptable solution}} \\ (3,2,4) & \Rightarrow (x,y,z) = \left(3,1,1 \right) & \color{#3D99F6}{\text{acceptable solution}} \\ (3,4,2) & \Rightarrow (x,y,z) = \left(3,2,\frac{1}{2} \right) & \color{#3D99F6}{\text{acceptable solution}} \\ (4,2,3) & \Rightarrow (x,y,z) = \left(4,1,\frac{3}{4} \right) & \color{#3D99F6}{\text{acceptable solution}} \\ (4,3,2) & \Rightarrow (x,y,z) = \left(4,\frac{3}{2},\frac{1}{2} \right) & \color{#D61F06}{\text{unacceptable solution}} \end{cases}$

Therefore, there are $\color{#3D99F6}{\boxed{5}}$ acceptable solution.