With ideal amplifiers and the feedback arrangements shown, you may deduce that the voltage on the [right hand] R1 node to the right of V3 must be V2 volts.
So the conventional current [from left to right] in that R1 must be
I=(V3-V2)/R1 amps.
That same current I flows in the feedback R2 connected to the output node. The left hand end of that same feedback R2 is (as previously noted) at voltage V2 volts.
From these relationships and your knowledge of V3 in terms of V1 you should be able to deduce Vo.
You can also treat this as two separate amplifiers.
v3 is the output of the first non-inverting amp due to v1.
The second amp has both a non-inverting output portion from v2 and an inverting output portion from v3. You calculate the two values separately and then add then together by superposition (noting that the inverting output polarity is subtracted from the non-inverting output polarity) to get v0.
It's important (I would even say rather critical) to note that if you can get V3 in terms of V1, then you should be able to get vV0 in terms of V1 and V2. After all, both V3 and V0 are the result of fundamentally identical circuits.
So the real question this raises is, when you say you can get V3 in terms of V1, how are you able to do this? Are you simply applying a formula from someplace, or are you analyzing the circuit?
If you are analyzing the circuit, then note that at the far left you ground, or 0V. Instead of analyzing it with a hard and fast 0V there, analyze it with an arbitrary voltage there, say Vg, and once you are done, substitute in 0V for Vg. Now notice that, with this analysis complete, you have already analyze the second part of the circuit and need merely replace Vg for that part with V3 and replace V1 with V2 and you are done.