Define $f(x,y,z)=\frac{(xy+yz+zx)(x+y+z)}{(x+y)(y+z)(z+x)}$.
Determine the set of real numbers $r$ for which there exists a triplet of positive real numbers satisfying $f(x,y,z)=r$.
1. Define the function \( f(x,y,z) = \frac{(xy+yz+zx)(x+y+z)}{(x+y)(y+z)(z+x)} \).

2. Using Vieta's formulas, we can express the symmetric sums in terms of the roots of a polynomial. Let \( x, y, z \) be the roots of the polynomial \( a^3 - a_1 a^2 + a_2 a - a_3 = (a-x)(a-y)(a-z) \). Here, \( a_1 = x+y+z \), \( a_2 = xy+yz+zx \), and \( a_3 = xyz \).

3. From \( x+y+z = a_1 \), we have:
   \[
   x+y = a_1 - z, \quad y+z = a_1 - x, \quad x+z = a_1 - y
   \]

4. Therefore, the product \((x+y)(y+z)(x+z)\) can be written as:
   \[
   (x+y)(y+z)(x+z) = (a_1 - x)(a_1 - y)(a_1 - z)
   \]

5. Expanding this product using the polynomial, we get:
   \[
   (a_1 - x)(a_1 - y)(a_1 - z) = a_1^3 - a_1^2(x+y+z) + a_1(xy+yz+zx) - xyz = a_1 a_2 - a_3
   \]

6. Substituting these into the function \( f(x,y,z) \), we have:
   \[
   f(x,y,z) = \frac{a_1 a_2}{a_1 a_2 - a_3} = 1 + \frac{a_3}{a_1 a_2 - a_3}
   \]

7. We need to determine the range of \( \frac{xyz}{(x+y)(y+z)(x+z)} \). Notice that:
   \[
   \frac{xyz}{(x+y)(y+z)(x+z)} \geq 0
   \]

8. To find the upper bound, consider the case when \( x = y = z \). Then:
   \[
   \frac{xyz}{(x+y)(y+z)(x+z)} = \frac{x^3}{(2x)(2x)(2x)} = \frac{x^3}{8x^3} = \frac{1}{8}
   \]

9. Therefore, we have:
   \[
   0 \leq \frac{xyz}{(x+y)(y+z)(x+z)} \leq \frac{1}{8}
   \]

10. Using the AM-GM inequality, we can show that:
    \[
    (x+y)(y+z)(x+z) \geq 8xyz
    \]
    This confirms that \( \frac{xyz}{(x+y)(y+z)(x+z)} \leq \frac{1}{8} \).

11. To show that \( \frac{xyz}{(x+y)(y+z)(x+z)} \) can take any value in the interval \( [0, \frac{1}{8}] \), let \( y = z = 1 \). Then:
    \[
    \frac{xyz}{(x+y)(y+z)(x+z)} = \frac{x}{2(x+1)^2}
    \]

12. Suppose \( b \) is a value such that \( 0 \leq b \leq \frac{1}{8} \). We need to find \( x \) such that:
    \[
    \frac{x}{2(x+1)^2} = b
    \]

13. Rearranging, we get the quadratic equation:
    \[
    2bx^2 + (4b-1)x + 2b = 0
    \]

14. Solving this quadratic equation using the quadratic formula:
    \[
    x = \frac{(1-4b) \pm \sqrt{(1-4b)^2 - 4 \cdot 2b \cdot 2b}}{4b} = \frac{(1-4b) \pm \sqrt{1-8b}}{4b}
    \]

15. Since \( b \leq \frac{1}{8} \), the discriminant \( 1-8b \geq 0 \), ensuring that \( x \) is a real value. Taking the positive root:
    \[
    x = \frac{(1-4b) + \sqrt{1-8b}}{4b}
    \]

16. Therefore, for \( y = z = 1 \) and \( x \) as defined above, \( \frac{xyz}{(x+y)(y+z)(x+z)} \) can take any value in \( [0, \frac{1}{8}] \).

17. Adding 1 to this range, the range of \( f(x,y,z) \) is \( [1, \frac{9}{8}] \).

The final answer is \( \boxed{ [1, \frac{9}{8}] } \)