First, you will need to recognize this is a redox problem. Once you know that, determine which species is being reduced and which one is being oxidized. This is quickly accomplished by writing the oxidation numbers above each atom.
Once you do that, you will see that iodine is being both reduced and oxidized. The reduction occurs because it is going from +5 in iodate to -1/3 in triiodide. The oxidation occurs because it is going from -1 in iodide to -1/3 in triiodide.
For iodine in iodate to go from +5 to -1/3, it has a change of -5 1/3 (or -16/3).
For iodine in iodide to go from -1 to -1/3, it has a change of +2/3.
You can now see that we need 8 iodides for every 1 iodate (2/3 x 8 = 16/3). The rest balances easily after that.
In this case, balancing by inspection is easier than redox. Start by looking at the Os and the Hs. On the left side, you need 1 IO3- to go with 6 H+ in order to make H2O.
1 IO3- + x I- + 6 H+ ---> y I3- + 3 H20
Ignoring I- and I3-, the total charge on the left side is +5 and the total charge on the right side is -1, so we are off by six charges and need to make that up by balancing I- and I3-. y = 3x -1, so I started plugging in x and y combos until the charge was balance. y=2 and x=5 gives 0 on the left and -2 on the right. y=3 and x=8 gives -3 on the left and -3 on the right.