Label the couples A, B, C so that the six people are A$_{1}$, A$_{2}$, B$_{1}$, B$_{2}$, C$_{1}$ and $C_{2}$.
Now, consider your 6 seats: _ _ _ _ _ _
There are 6 choices for who can sit in the first seat; without loss of generality, say it's A$_{1}$.
There are 4 choices for who can sit in the second seat (anyone else but A$_{2}$); without loss of generality, let's say this is B$_{1}$.
Our table now looks like this: A$_{1}$ B$_{1}$ _ _ _ _
For the 3rd seat, we just can't have B$_{2}$, so we can either have A$_{2}$ or one of the C's; I'll split this into 2 cases.
Case 1: A$_{2}$ sits in the third seat.
A$_{1}$ B$_{1}$ A$_{2}$ _ _ _
Now the fourth seat can't be B$_{2}$, otherwise the C's will sit next to each other in the last two seats. So the fourth seat must be one of the C's; there are 2 choices for this. Then the fifth seat must be B$_{2}$ and the sixth must be the other C.
Thus there are $6*4*1*2*1*1 = 48$ ways for Case 1 to happen.
Case 2: One of the C's sits in the third seat; there are 2 possibilities for this. Let's say it's C$_{1}$
A$_{1}$ B$_{1}$ C$_{1}$ _ _ _
Then there are 2 choices for the fourth seat: A$_{2}$ or B$_{2}$. I'll treat these as subcases.
Case 2.1: A$_{2}$ sits fourth.
A$_{1}$ B$_{1}$ C$_{1}$ A$_{2}$ _ _
For the fifth seat, we have 2 choices, and then 1 for the sixth.
Thus $6*4*2*1*2*1 = 96$ possibilities for this case.
Case 2.2: B$_{2}$ sits fourth.
A$_{1}$ B$_{1}$ C$_{1}$ B$_{2}$ _ _
For the fifth seat, there is only one possibility: A$_{2}$, since otherwise A$_{2}$ would be in the sixth seat, which is adjacent to the first (the table is circular). Of course there is only one possibility for the sixth.
Thus there are $6*4*2*1*1*1 = 48$ ways for Case 2.2 to happen.
There are then $48+96+48 = 192$ possible seating arrangements. The probability of this happening is $\frac{192}{6!} = \frac{192}{720} = \frac{4}{15} \approx 0.267$
You could do it this way.
- Couple sits at the end, so the arrangement must be $X_1X_2Y_1Z_1Y_2Z_2$. The number of choices for the six people, going left to right, are $6,1,4,2,1,1$, total $48$.
- One away from the end: $Y_1X_1X_2Z_1Y_2Z_2$, choices $6,4,1,2,1,1$, total $48$.
- Sitting in the middle: $Y_1Z_1X_1X_2(any)(any)$, choices $6,4,2,1,2,1$, total $96$.
- The reverse of the first two cases, total $96$.
This gives your $288$.
If you wanted, you could instead do the whole thing by inclusion/exclusion. Let $A_1$ be the set of arrangements in which the first couple sit together, $A_2$ and $A_3$ likewise. Then the number we need is
$$\eqalign{|\overline A_1\cap\overline A_2\cap\overline A_3|
&=|{\cal U}|-|A_1\cup A_2\cup A_3|\cr
&=|{\cal U}|-|A_1|-|A_2|-|A_3|+|A_1\cap A_2|+|A_1\cap A_3|\cr
&\qquad\qquad\qquad\qquad\qquad\qquad{}+|A_2\cap A_3|-|A_1\cap A_2\cap A_3|\ .\cr}$$
The total number of arrangements of $6$ people is $6!\,$. To count $A_1$,
- pick the places for the first couple to sit: as they are sitting together, this amounts to arranging the letters $PPPPC$, where $P$ is a person and $C$ is a couple. . . . . there are $5$ ways to do this;
- pick which way round the members of the couple sit. . . . . $2$ ways;
- order the other $4$ people. . . . . $4!$ ways.
So $|A_1|=5\times2\times4!\,$, and clearly $|A_2|,|A_3|$ are the same. For similar reasons, starting by ordering $PPC_1C_2$ we have
$$|A_1\cap A_2|=P(4,2)\times2^2\times2!\ ,\quad
|A_1\cap A_2\cap A_3|=3!\times2^3$$
and so our answer is
$$6!-3\times5\times2\times4!+3\times P(4,2)\times2^2\times2!-3!\times2^3
=6!-6!+288-48=240\ .$$
Best Answer
For $A_2$ you have a good idea, but you missed a detail. There are $3$ ways to choose which couple does not sit together; this is the factor that you missed. Then there are $3$ ways to split that couple (your $P_1aP_2b$, $aP_1bP_2$, and $aP_1P_2b$), $2$ ways to decide which couple is $P_1$ and which is $P_2$, and $2$ ways to seat the members of each couple, so
$$A_2=3\cdot3\cdot2\cdot2^3=9\cdot16=144\;.$$
In terms of your calculation, that changes $4!-2!\cdot3$ to $4!-3\cdot2!\cdot3=6$, and $2!^3\cdot3\cdot6=144$.
Probably the easiest way to calculate $A_3$ is to subtract the total of the other three possibilities from $6!$, since you already have them.
Added: Since there are only three couples, $A_4$ isn’t too hard to calculate by hand. Let $A$ be the person in the first seat in the row; there are $6$ choices for $A$. Let $A'$ be $A$’s spouse. Anyone except $A'$ can sit in the second seat, so there are $4$ possibilities; call the person who sits there $B$. Now split the count into two cases.
The third seat is taken by $C$, a member of the third couple. There are $2$ ways to choose $C$, so there are altogether $6\cdot4\cdot2$ ways to fill the first three seats with members of three distinct couples. The fourth seat can be occupied by $A'$ or $B\,'$, and in either case the last two people can sit in either order, so there are $2^2=4$ ways to fill out the row, for a total of $6\cdot4\cdot2\cdot4=192$ arrangements.
The third seat is taken by $A'$, so that we have $ABA'$ in the first three seats. The rest of this calculation is spoiler-protected to give you a chance to finish it on your own.