Let's say we want to count $3 \times 9$ Sudoku grids, i.e. grids whose entries are taken from $\{1,\dots,9\}$ and such that no rows or $3 \times 3$ sub-grids contain repetitions.
\begin{matrix} 1 & 2 & 3 && 4 & 5 & 6 && 7 & 8 & 9 \\ 7 & 8 & 9 && 1 & 2 & 3 && 4 & 5 & 6 \\ 4 & 5 & 6 && 7 & 8 & 9 && 1 & 2 & 3 \end{matrix}
My strategy was the following one. There are $9!$ distributions of values for the first row. I'm assuming that if we settle values for the first and second rows, the third one is fixed. (How can I prove this?). Hence, we only have to determine the values for the second row.
Notice that if we replace each of the three triples that make each row with variables $A_1,\dots,B_2,\dots$, etc., we get:
\begin{matrix} A_1 && A_2 && A_3 \\ B_1 && B_2 && B_3 \\ C_1 && C_2 && C_3 \end{matrix}
Now we can clearly see that $B_2$ can't share values with $A_2$. Hence, $A_2$ must distribute its values to $B_1$ and $B_3$ in a symmetric manner: if $B_1$ gets two values from $A_2$ and one from $A_3$, $B_3$ gets one value from $A_2$ and two from $A_1$, etc. If we choose values for $B_1$ and $B_3$ in this way, we're effectively also fixing $B_2$'s values, so we can then count their permutations.
Is $(1)$ the final answer?
$$9! \bigg[ \sum_{k=0}^3 \bigg( {3\choose k}{3 \choose 3-k}3! \bigg)^2 3! \bigg] \tag 1$$
$$9! \bigg[ \sum_{k=0}^3 \bigg( {3\choose k}{3 \choose 3-k}3! \bigg)^2 3! \bigg] \tag 1$$
is not correct. The point that the number of numbers in the first three of the second row that are taken from the second three in the first row matches the number in the second three taken from the third three is correct. I would pull the $3!$ out of the sum, because the sum allocates the numbers to the groups of three in the second row. You then have $3!^3$ because you can reorder each group of three as you wish. That gives your expression, but I think makes it more clear how you found it. Then in the third row the set of numbers in each group of three is determined but they can come in any order, which gives another factor of $3!^3$ so I would represent it as $$9! 3!^6\bigg[ \sum_{k=0}^3 \bigg( {3\choose k}{3 \choose 3-k} \bigg)^2 \bigg] \tag 1$$