Where is my solution wrong for this combinatorics problem?

You are overcounting the arrangements. For example, by choosing three $X$s in the three rows and then filling the remaining places with three $x$s, we consider the following equivalent arrangements as different

 Xo    Xo    Xo    Xo
Xxxx  xXxx  xxXx  xxxX 
 Xo    Xo    Xo    Xo

Note that we have $\binom{8}{6}$ ways to place six $X$s in the eight cells. From this number we subtract $2$, the ways to have a row empty (the first and the third): $$\binom{8}{6}-2=\frac{8\cdot 7}{2}-2=28-2=26.$$

The problem with your solution is that you count a lot of the combinations twice. For instance, if we first add

 xo
xooo
 xo

and then add the rest of the three to form

 xo
xxxx
 xo

it is the same as starting with

 xo
ooxo
 xo

and then adding the rest of the x to form

 xo
xxxx
 xo

Now instead of doing the difficult thing and removing the duplicates of your solution do it like this: Just place all x. This can be done in $8 \choose 6$ ways. However when we fill the top row and the middle row, or the bottom row and the middle row, there is one row without an x. Thus we have to subtract $2$. Hence the solution is ${8 \choose 6 } - 2 = 26$.