What is the best approach to balance this equation?
$$\ce{CO2 + H2O <--> C2H2 + O2}$$
I did it at first this way and I was not successful: \begin{align} \ce{4CO2 + H2O &<--> C2H2 + 3O2}\\ \ce{2CO2 + 6H2O &<--> 6C2H2 + 5O2}\\ \ce{12CO2 + 6H2O &<--> 6C2H2 + 5O2}\\ \end{align}
Should I start from right side or left? They say start from the most complicated molecule? If it is like that, then what is the most complicated one and why?
This may be a bit dumb method, but sometimes I do use it as it is pretty straightforward and doesn't require any guessing. I setup a few equations and try solving over integers. $$\ce{aCO2 + bH2O <--> cC2H2 + dO2}$$
Balancing $\ce{C}$ atoms both sides, $$a = 2c\\$$ Balancing $\ce{H}$ atoms both sides, $$2b=2c\\$$ Balancing $\ce{O}$ atoms both sides,
$$2a+b=2d$$
Solving gives this. Not hard to see that $a=4$ does the job.
The key for this type of equation is to balance the diatomic molecule last, and then just add a 1/2 of the diatomic molecule to finish the balance. Then, double all the coefficients and you're done.
balance everything but the diatomic molecule: $$\ce{2CO2 + H2O <=> C2H2 + O2}$$
Then add the right fraction of the diatomic molecule: $$\ce{2CO2 + H2O <=> C2H2 + 2.5O2}$$
Notice that 2.5 oxygen molecules is 5 oxygen atoms.
There is a standard equation for balancing such reactions: $$\ce{C_xH_y + [x +(y/4)]O2 -> x CO2 + y/2 H2O}$$
Hopefully, now you should be able to apply it to your reaction. It's just the reverse reaction that you're dealing with.
