This is a pretty tricky problem, since there are multiple atoms and it isn't immediately clear what is getting oxidized and reduced. To figure that out, we need to look at the oxidation states of the individual atoms.
First let's look at VO 2+. V is a transition metal, which means we have to figure out its oxidation state by context. O is most electronegative, so we assign it its usual oxidation state (-2). Overall charge on VO is 2+, so the V must balance to provide our overall charge - V's oxidation state must be +4.
Now let's take a look at VO2 + on the other side. We use the same technique to figure out oxidation states: O's -2. Since there are two O atoms, O is contributing -4 to the overall oxidation state. Since the overall oxidation state is +1, we know that V's oxidation state must balance that out accordingly. So V's oxidation state is +5.
So what happened to the oxidation state of the individual atoms as we went from VO 2+ to VO2 +?
O's oxidation state: -2 to -2 (no change)
V's oxidation state: +4 to +5 (lost 1 electron)
O's didn't lose or gain electrons, so it wasn't reduced here. V lost an electron. Therefore VO 2+ couldn't be possibly be the oxidizing agent, and in fact must be the reducing agent.
So if VO 2+ didn't act as the oxidizing agent, what did? Taking a look at our reactants, we see O2 (g) in its standard state, which means the O atoms have an oxidation state of 0. The O's on the product side have an oxidation state of -2, so it fits that O2 acted as the oxidizing agent and was itself reduced (gained electrons). Therefore O2 is the oxidizing agent, and VO 2+ is the reducing agent.