I would generally start looking at forced chains of cells with 2 options left.
In this particular case my first idea would be to follow the chain
A3 -> A6 -> B6 -> B1 -> C2 -> C3
either A3 is a 2 or if it's a 6 (following the forced chain) C3 is a 2.
Thus B4 and F3 cannot be 2.
And next F needs a 2 in top block so D2 can't be a 2
Similar chaing reasoning can be used to see if a cell with 3 options is part of such a chain as well to reduce it to only 2 options instead:
In this case (after previous steps) I would look at F2 starting the chain from F3:
F3 = 4 -> F2 = 4
F3 = 6 -> A3 = 2 -> C3 = 4 -> C2 = 2 -> F2 = 6
=> F2 cannot be 2
Which then leads us to F1 = 2 and solving the rest trivially.
I do want to point out though that the above steps took longer than your trial and error solution.
Which one is the best way to solve it depends largely on what you want.
I personally don't like guessing without reasoning so will only fall back on trial and error if all other options are exhausted ... or at least untill it takes too long to spot any of those other options. Nothing wrong with at least trying something instead of staring at a really hard puzzle without progress.
Final note: sometimes when just trying something randomly makes it easier to spot other forced hints even if your initial guess didn't directly lead to reducing any options.