Now that you've gotten the answer, here is a little bit of a post mortem clarifying the essential points, as may come up again in future problems. (In general, once you've solved a problem, it's a really good idea to ask for the "moral(s)". If you can't identify one, how are you better off now than before you solved it?)
In this case, I think the most important single step is the recognition that the conclusion
"$f(x) + g(x) = 6$ for all $x$"
can be more usefully rephrased as
"$f(x) + g(x)$ is a constant function, with constant value $6$."
This is much more suggestive, because a function $f$ defined on an interval is constant if and only if its dervative is constantly equal to zero. (Stationarity is equivalent to identically zero velocity.) Note that half of this statement is obvious, but the other half is not: it is consequence of the Mean Value Theorem that should be emphasized both in the text and by your instructor.
Thus you are clued in to the fact that the main thing you want to show is that $(f+g)' = f' + g' \equiv 0$ (i.e., constantly zero). Since a constant function is determined by plugging in any point, once you know that $f+g$ is constant, seeing that it's constantly equal to SOMETHING -- in this case $6$ -- shouldn't be a problem. So now you look and see how to get from what you're given to the conclusion that $f' + g' \equiv 0$, and you see that you're being given an expression for $f'$ and $g'$ separately. So certainly you want to add them together and hope to get $0$; in this case, that hope is immediately fulfilled.