In how many ways can three pairs of siblings from different families be seated in two rows of three chairs, if siblings may sit next to each other in the same row, but no child may sit directly in front of their sibling?
There are 6 possible configurations for the two rows of chairs, listed below:
1) ABC
DEF
2) ABD
CEF
3) ABE
CDF
4) ACF
BDE
5) ACE
BDF
6) ADE
BCF
For each configuration, we need to determine how many ways the siblings can be seated.
In the first configuration (ABC and DEF), there are 3 siblings in each row. We can arrange them in $3!$ ways in the first row and $3!$ ways in the second row. Therefore, there are $3! \times 3! = 36$ ways to arrange the siblings in this configuration.
In the second configuration (ABD and CEF), the first chair in the second row (D) cannot be occupied by child B, as this would violate the condition that no child may sit directly in front of their sibling. So, there are only 2 options for child B. The remaining siblings can be arranged in $2!$ ways in the first row and $3!$ ways in the second row. Therefore, there are $2! \times 3! \times 2 = 24$ ways to arrange the siblings in this configuration.
In the third configuration (ABE and CDF), the first chair in the second row (D) cannot be occupied by child B, and the second chair in the second row (E) cannot be occupied by either child B or child C. So, there is only 1 option for child B and 1 option for child C. The remaining siblings can be arranged in $1!$ way in the first row and $3!$ ways in the second row. Therefore, there are $1! \times 3! \times 1 = 6$ ways to arrange the siblings in this configuration.
For the remaining configurations (ACF and BDE, ACE and BDF, ADE and BCF), the same logic applies. There is only 1 option for certain children to occupy certain chairs, and the remaining siblings can be arranged in the remaining spaces.
Therefore, the total number of ways the siblings can be seated in two rows of three chairs is $36+24+6+6+6+6 = \boxed{84}.$