If positive real numbers $x,y,z$ satisfy the following system of equations, compute $x+y+z$.

$$xy+yz = 30$$
$$yz+zx = 36$$
$$zx+xy = 42$$

[i]Proposed by Nathan Xiong[/i]
1. We start with the given system of equations:
   \[
   xy + yz = 30
   \]
   \[
   yz + zx = 36
   \]
   \[
   zx + xy = 42
   \]

2. Add all three equations:
   \[
   (xy + yz) + (yz + zx) + (zx + xy) = 30 + 36 + 42
   \]
   Simplifying the left-hand side, we get:
   \[
   2(xy + yz + zx) = 108
   \]
   Divide both sides by 2:
   \[
   xy + yz + zx = 54
   \]

3. Now we have the equation:
   \[
   xy + yz + zx = 54
   \]

4. To find the individual products, we subtract each of the original equations from this sum:
   \[
   zx = 54 - (xy + yz) = 54 - 30 = 24
   \]
   \[
   xy = 54 - (yz + zx) = 54 - 36 = 18
   \]
   \[
   yz = 54 - (zx + xy) = 54 - 42 = 12
   \]

5. We now have:
   \[
   zx = 24, \quad xy = 18, \quad yz = 12
   \]

6. To find \(x, y, z\), we multiply these three equations:
   \[
   (zx)(xy)(yz) = 24 \cdot 18 \cdot 12
   \]
   Simplifying the right-hand side:
   \[
   (xyz)^2 = 24 \cdot 18 \cdot 12
   \]
   Calculate the product:
   \[
   24 \cdot 18 = 432
   \]
   \[
   432 \cdot 12 = 5184
   \]
   Thus:
   \[
   (xyz)^2 = 5184
   \]
   Taking the square root of both sides:
   \[
   xyz = \sqrt{5184} = 72
   \]

7. Now we solve for \(x, y, z\) using the individual products:
   \[
   yz = 12 \implies z = \frac{12}{y}
   \]
   \[
   zx = 24 \implies x = \frac{24}{z}
   \]
   \[
   xy = 18 \implies y = \frac{18}{x}
   \]

8. Substitute \(z = \frac{12}{y}\) into \(x = \frac{24}{z}\):
   \[
   x = \frac{24}{\frac{12}{y}} = \frac{24y}{12} = 2y
   \]

9. Substitute \(x = 2y\) into \(y = \frac{18}{x}\):
   \[
   y = \frac{18}{2y} \implies y^2 = 9 \implies y = 3
   \]

10. Using \(y = 3\), find \(z\) and \(x\):
    \[
    z = \frac{12}{y} = \frac{12}{3} = 4
    \]
    \[
    x = 2y = 2 \cdot 3 = 6
    \]

11. Therefore, \(x + y + z = 6 + 3 + 4 = 13\).

The final answer is \(\boxed{13}\)