See attached file
I'll assume that we are using units such that c=1. The two world lines are given by:
x1 = t/3
x2 = t^2/18
Here the "18" must carry the units of time used. So, we'll just assume that some units are chosen and the numbers that come out for times/distances are given in terms of that unit.
The time when the observers meet follows from equating x1 to x2:
x1 = x2 -->
t = 0 and t = 6.
At t = 6, x1 = x2 = 2.
The event where they meet again is thus the event t = 6, x = 2.
Now, you know from elementary special relativity that the relation between proper time tau and coordinate time t is:
tau = t/gamma
where gamma = 1/sqrt[1-v^2]
But you must be very careful about applying equations from special relativity derived for inertial observers to accelerating observers. So, let's apply this equation to observer 1 and then I'll show how you deal with the general case. Observer 1 moves ...
A detailed solution is given.