If two random customers pick four chopsticks at random, what is the probability that they pick one for each color?

Yes this is correct.

Here is another way to do it.

The first customer picks a chopstick from the set of eight, this will always be only one color so the "distinct color" probability for this pick is $8/8=1$. Then he picks the second chopstick, this time six out of the remaiming seven obey the distinct color rule so this gives a factor of $6/7$. For the second customer his first chopstick gives a factor of $4/6$ (four "good" choices out of six total) and then $2/5$ for the last selection. So we have the overall probability

$\dfrac{8×6×4×2}{8×7×6×5}=\dfrac{8}{35}$

as claimed.