I think your idea is basically right. Remember, however, that there are two types of paths that have been double-counted: paths that pass through a corner adjacent to the initial corner, and paths that pass through a corner adjacent to the final corner. Therefore you need a factor of $6$ rather than $3$ in your second term.
Here's an inclusion-exclusion argument that clarifies things—at least for me: let the ant start at the front, bottom, left corner, and end at the back, top, right corner. Let the faces adjacent to the initial corner be called $F$ (front), $B$ (bottom), and $L$ (left), and let the faces adjacent to the final corner be called $F'$ (back), $B'$ (top), and $L'$ (right).
As you say, the ant's path lies within two faces. One of these will be an unprimed face; one will be a primed face. Define $P(X,X')$ to be the set of paths that lie within faces $X$ and $X'.$ We have
$$
\lvert P(X,X')\rvert=\begin{cases}0 & \text{if $X$ and $X'$ are opposite each other,}\\
\binom{9}{3} & \text{otherwise.}\end{cases}
$$
To use an inclusion-exclusion argument, we need to know the sizes of intersections of sets $P(X,X').$ As stated in your solution, intersections like $P(B,F')\cap P(L,F')$ contain $\binom{6}{3}$ paths since such paths must traverse the edge where $B$ and $L$ meet, leaving $\binom{6}{3}$ ways to get from one corner of $F'$ to the other. The same is true of intersections like $P(F,B')\cap P(F,L').$ There are six such intersections in all. The other non-empty two-way intersections are sets like $P(F,B')\cap P(B,L'),$ which contains the single path that traverses the edge where $F$ and $B$ meet, the edge where $F$ and $L'$ meet, and the edge where $B'$ and $L'$ meet. There are six of these as well. Finally, the only non-empty three-way intersections are sets like $P(F,B')\cap P(F,L')\cap P(B,L'),$ which contains the single path described above. There are six of these too.
Inclusion-exclusion then says that the number of paths is the sum of the sizes of all the sets $P(X,X')$ minus the sum of the sizes of all two-way intersections plus the sum of the sizes of all three-way intersections. This yields
$$
6\cdot\binom{9}{3}-\left(6\cdot\binom{6}{3}+6\cdot1\right)+6\cdot1=6\cdot\binom{9}{3}-6\cdot\binom{6}{3}=384.
$$
As the hint suggested, to reach $(5,5)$ from $(0,0)$, we will take $10$ consecutive "steps," of which $5$ will be up and $5$ to the right. We can choose any $5$ of these $10$ steps to be the "up" steps.
So there are $\binom{10}{5}$ possible paths.
Best Answer
You have to move over the surface of the 2-unit cube; you're not allowed to go through the centre of the cube.