First, judging from your work, you misstated the question: it appears that you meant to ask for the probability that no husband sits next to his wife. It also appears that the seats are arranged in a row, not a circle. I will make these assumptions.
Your first three factors, $8\cdot6\cdot5$, in the numerator are fine. After that, though, you have to split the calculation into cases. Suppose that the first three people, in order, are A, B, and C. If A and C are a couple, then any of the remaining $5$ people can sit in the fourth seat. Your calculation, with a factor of $4$, is correct only if C’s spouse has not already been seated, i.e., C and A are not a couple. And the cases just proliferate after that, which is why you’re better off working with the complement.
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 $\frac{192}{6!} = \frac{192}{720} = \frac{4}{15} \approx 0.267$
Best Answer
Just see how you can have no couples sitting together.
So let's say we have $112233$ that we need to mix up so that no two of the same are next to each other.
Start with two different ones:
$$12$$
The next can be a $1$, and then the rest is forced: $121323$
The next can also be a $3$, and then you have a few options: $123123, 123132, 123213, 123231$
OK, so that's $5$ options, but since the first two can be $12, 21, 13, 31, 23,$ or $32$ you get $6 \cdot5 = 30$ ways for this to happen .... out of $\frac{6!}{2!2!2!}=90$ total ways... giving you a probability of $$\frac{30}{90}=\frac{1}{3}$$