Here is the question I am trying to solve:
Compute $S(E^i F^j K^l)$ in $U_q.$
Here is my thoughts:
Definition: We define $U_q = U_q(\mathfrak{sl}(2))$ as the algebra generated by the four variables $E , F, K, K^{-1}$ with the relations:
$$ \begin{align} KK^{-1} &= K^{-1}K = 1\\ KEK^{-1} &= q^2 E \\ KFK^{-1} &= q^{-2}F \\ [E,F] &= \frac{K - K^{-1}}{q - q^{-1}} \end{align} $$
I also know the following relations where $S$ is the antipode:
$$S(E) = -EK^{-1} \quad S(F) = -KF \quad S(K) = K^{-1} \quad S(K^{-1}) = K.$$
Now, my trial is as follows:
$S(E^i F^j K^l) = S(K^l) S(F^j) S(E^i) = (S(K))^l (S(F))^j (S(E))^i = (K^{-1})^l (-KF)^j (-EK^{-1})^i$ where the first equality because $S$ is antihomomorphism and the second equality by the properties of antihomomorphism and the third equality by some of the properties given above.
But then I do not know how to complete, can anyone help me in this please?
These computations can get somewhat complicated, and eventually it's worth getting a computer algebra system like sage to do it for you (see the documentation here, for instance). Of course, it's always a good idea to build intuition by doing these sorts of computations by hand, so let's do a special case together. Hopefully it will be clear how to extend this to general $(i,j,k)$ -- I just think the exposition will be clearer without needing to manage the indices.
As one extra caveat, I'm not sure if it's a difference in convention or a typo, but I think the antipode equations for $E$ and $F$ should be
$$S(E) = - K^{-1} E \quad S(F) = -FK$$
This agrees with both sage and most of the references I'm able to quickly look at. Either way, it should be clear how to modify the computation in case this is a convention difference rather than a typo.
So then let's compute
$$S(E^2 F^3 K)$$
This is big enough to be interesting, but small enough to be manageable. As you've said, since $S$ is an antihomomorphism, this becomes
$$S(K) \cdot S(F)^3 \cdot S(E)^2$$
which has $S$s to get rid of
$$K^{-1} \cdot (-FK)^3 \cdot (-K^{-1}E)^2$$
Now we can pull out the negative signs to the front, and fully expand this to get
$$- (K^{-1}) (FK \ FK \ FK)(K^{-1} E \ K^{-1} E)$$
which, after cancelling the $KK^{-1}$ in the middle, becomes
$$- K^{-1} \ FK \ FK \ F E \ K^{-1} E$$
Now to finish things off, we just need to push all the $E$s to the left and all the $K$s to the right! The key here is to rephrase our conjugation identities. For instance, we rearrange $KEK^{-1} = q^2 E$ to the more helpful $K^{-1} E = q^{-2} E K^{-1}$. This lets us move a $K^{-1}$ to the right of an $E$, as long as we remember to pick up some power of $q$.
In particular, this lets us rewrite the above term as
$$-q^{-2} \ K^{-1} \ FK \ FK \ F E^2 K^{-1}$$
Now, we could keep doing this, and slowly move our $E$s to the left and our $K$s to the right. But there's some shortcuts we can take too! For instance, we recognize a $K^{-1}FK$ on the left. Rewriting $KFK^{-1} = q^{-2} F$ as $K^{-1} F K = q^2 F$ lets us turn our term into
$$ - q^{-2} \ q^2 F \ FK \ F E^2 K^{-1} = - F^2 K F E^2 K^{-1}$$
At this point, you probably get the game. We'll move the $K$ past the $F$ by rewriting $KFK^{-1} = q^{-2} F$ as $KF = q^{-2} FK$. This brings us to
$$ -q^{-2} \ F^3 \ K E^2 K^{-1} $$
Now, again, we could brute-force this $K$ to the right of the $E^2$, but we notice another opportunity for a trick! We notice $K E^2 K^{-1} = (K E K^{-1})^2 = (q^2 E)^2$. So we might as well save some work by using this!
$$ - q^2 \ F^3 E^2 $$
This is probably a good place to stop. Of course, you might want to write this as a sum of monomials of the form $E^i F^j K^l$, and in that case all that's left is to push the $E$s past the $F$s. I won't show this in full (you'll see why after the break) but I'll at least show you what the process looks like, and you can finish yourself if you're so inclined.
All that's left is to push the $E$s past the $F$s. But we know that $EF - FE = [E,F] = \frac{K - K^{-1}}{q - q^{-1}}$. Or, said another way, $FE = EF - \frac{K - K^{-1}}{q - q^{-1}}$! This brings us to the most painful part of the computation:
$$- q^2 \ F^2 \ FE \ E$$
becomes
$$-q^2 \ F^2 \left ( EF - \frac{K - K^{-1}}{q - q^{-1}} \right ) \ E$$
or
$$-q^2 \ F^2 \ EF \ E + \frac{q^2}{q - q^{-1}} \ F^2 K E - \frac{q^2}{q - q^{-1}} F K^{-1} E$$
Now, unfortunately, the process continues. And each time we push an $E$ past an $F$ we pick up more terms. This is why I suggest using a computer, haha. Hopefully it's clear by now how the rest of this computation will go. We keep moving $K$s to the right and $E$s to the left, picking up factors of $q$ whenever we push a $K$ past something, and picking up a new summand whenever we move an $E$ past an $F$.
I'll leave it as a (quite annoying) exercise to finish this expansion if the mood strikes you. I'll also say that the case of general computations $S(E^i F^j K^l)$ proceed exactly like this! As you might guess, though, managing the indices in the abstract is even more annoying than managing the concrete case of $(2,3,1)$ that we did here. It's possible in principle to get an honest closed form formula, but it would necessarily be somewhat nasty. Imo it's much better to get experience working these out in enough special cases that you know roughly how the computation proceeds, which is what I've tried to emphasize here.
I hope this helps ^_^