Determine all positive integer solutions $(x, y, z, t)$ of the equation \[(x+y)(y+z)(z+x)=xyzt\] for which $\gcd(x, y)=\gcd(y, z)=\gcd(z, x)=1$.
1. Given the equation \((x+y)(y+z)(z+x) = xyzt\) and the conditions \(\gcd(x, y) = \gcd(y, z) = \gcd(z, x) = 1\), we start by analyzing the equation. 

2. We introduce a new variable \( k \) such that:
   \[
   k = \frac{x+y+z}{xyz}
   \]
   Since \( k \) must be a positive integer, we can write:
   \[
   x + y + z = k \cdot xyz
   \]

3. Without loss of generality, assume \( x \leq y \leq z \). This implies:
   \[
   x + y + z \leq 3z
   \]
   Substituting \( x + y + z \) from the previous equation, we get:
   \[
   k \cdot xyz \leq 3z
   \]
   Dividing both sides by \( z \):
   \[
   k \cdot xy \leq 3
   \]
   Therefore:
   \[
   xy \leq \frac{3}{k}
   \]

4. Since \( k \) is a positive integer, the possible values for \( xy \) are limited. We consider the cases where \( xy \leq 3 \).

5. **Case 1: \( (x, y) = (1, 1) \)**
   \[
   k = \frac{2 + z}{z}
   \]
   For \( k \) to be an integer, \( 2 + z \) must be divisible by \( z \). This implies:
   \[
   2 + z = kz
   \]
   Solving for \( z \):
   \[
   z(k - 1) = 2 \implies z = \frac{2}{k-1}
   \]
   The possible values of \( k \) are 2 and 3:
   - For \( k = 2 \):
     \[
     z = \frac{2}{2-1} = 2
     \]
     This gives the solution \((x, y, z) = (1, 1, 2)\).
   - For \( k = 3 \):
     \[
     z = \frac{2}{3-1} = 1
     \]
     This gives the solution \((x, y, z) = (1, 1, 1)\).

6. **Case 2: \( (x, y) = (1, 2) \)**
   \[
   k = \frac{3 + z}{2z}
   \]
   For \( k \) to be an integer, \( 3 + z \) must be divisible by \( 2z \). This implies:
   \[
   3 + z = 2kz
   \]
   Solving for \( z \):
   \[
   z(2k - 1) = 3 \implies z = \frac{3}{2k-1}
   \]
   The possible value of \( k \) is 1:
   - For \( k = 1 \):
     \[
     z = \frac{3}{2 \cdot 1 - 1} = 3
     \]
     This gives the solution \((x, y, z) = (1, 2, 3)\).

7. **Case 3: \( (x, y) = (1, 3) \)**
   \[
   k = \frac{4 + z}{3z}
   \]
   For \( k \) to be an integer, \( 4 + z \) must be divisible by \( 3z \). This implies:
   \[
   4 + z = 3kz
   \]
   Solving for \( z \):
   \[
   z(3k - 1) = 4 \implies z = \frac{4}{3k-1}
   \]
   The possible value of \( k \) is 1:
   - For \( k = 1 \):
     \[
     z = \frac{4}{3 \cdot 1 - 1} = 2
     \]
     This gives the solution \((x, y, z) = (1, 3, 2)\), but this does not satisfy \( y \leq z \).

8. **Case 4: \( (x, y) = (2, 2) \)**
   \[
   k = \frac{4 + z}{4z}
   \]
   For \( k \) to be an integer, \( 4 + z \) must be divisible by \( 4z \). This implies:
   \[
   4 + z = 4kz
   \]
   Solving for \( z \):
   \[
   z(4k - 1) = 4 \implies z = \frac{4}{4k-1}
   \]
   There are no integer solutions for \( z \).

9. **Case 5: \( (x, y) = (2, 3) \)**
   \[
   k = \frac{5 + z}{6z}
   \]
   For \( k \) to be an integer, \( 5 + z \) must be divisible by \( 6z \). This implies:
   \[
   5 + z = 6kz
   \]
   Solving for \( z \):
   \[
   z(6k - 1) = 5 \implies z = \frac{5}{6k-1}
   \]
   There are no integer solutions for \( z \).

10. **Case 6: \( (x, y) = (3, 3) \)**
    \[
    k = \frac{6 + z}{9z}
    \]
    For \( k \) to be an integer, \( 6 + z \) must be divisible by \( 9z \). This implies:
    \[
    6 + z = 9kz
    \]
    Solving for \( z \):
    \[
    z(9k - 1) = 6 \implies z = \frac{6}{9k-1}
    \]
    There are no integer solutions for \( z \).

Conclusion:
The only positive integer solutions \((x, y, z)\) that satisfy the given equation and conditions are:
\[
(x, y, z) = (1, 1, 1), (1, 1, 2), (1, 2, 3)
\]

The final answer is \( \boxed{ (1, 1, 1), (1, 1, 2), (1, 2, 3) } \).