May I connect 3.6v Li-pol accumulator to 3.3v dc supply?
There is a way to join two supplies: joining diodes.

Let me explain what's going on here:
A and B are your voltage sources. Imagine there is no R. So the Out will be powered from the higher of the two voltages (A-ddrop or B-ddrop), and the lower voltage source will give no current, it won't be able to send current to the other side of the diode, the voltage there is too high for our lower voltage source. Put a large cap instead of R, and if one of the supplies fails, the other one will keep powering the thing, and the diodes will block current in the reverse direction (if A shuts down, you don't want current from B to flow into A).
As for voltage levels - you'll have to consult the datasheet. Considering the clock thing uses very tiny current, you may just stupidly want to discharge Li-Po to 3.3V, it will still be enough for eternity, especially considering you will actually use its power very little, only during your blackouts. Most likely, you can't go over 3.3V with that thing (well, nobody will die from 3.4, but don't risk going higher). You have to look up the chip specs, but those things usually run only at 3.3. If you have no idea, assume 3.4V is maximum. So use two schottky diodes (available in ebay diode sets too), they have 0.15V voltage drop, so discharge Li-Po to 3.4 and hook it up. And yes, you may need to cut a trace or something to put a Schottky between 3.3VDC source and the clock (it will work fine from slightly dropped 3.15V voltage too).