The MLZ series of inductors is not designed for power conversion (buck, boost, etc.), but instead for power supply filtering. This means that a) the windings are not optimized for low skin losses, and b) the magnetic core is not optimized for low core loss. From a circuit design perspective, this means the boost regulator will be less efficient.
If efficiency is not what you're after -- if you simply care for functionality and compactness -- then I don't see any problem using MLZ (or other "ferrite choke") inductors, as long as your circuit doesn't overheat. In practice, this means sticking to low power levels. Just ensure the current rating covers your needs (with healthy margin), and evaluate the power supply thoroughly.
LTSpice has a model for the controller built in, so you can model the system fairly easily. This doesn't take into account effects from your actual PCB layout. While you could try and model those effects, it's best to just build the circuit go from there.
I generally start with the default values and go from there, unless there's some known stuff I can simulate which will let me adjust them.
I like to simulate converters with various kinds of loads. Simple switched loads like what's in the screenshot can give you some basic indications as to what will happen under certain conditions. You can also use arbitrary sources to force certain loads and see how the converter handles those, but they can cause odd situations to happen if you're not careful.
If you wind up with crazy voltage spikes, try lowering your timestep size.
This also lets you try a whole range of different configurations at once. This screenshot shows the output voltage for 10k steps in compensation resistor from 1k to 300k.
The datasheet actually shows how the system works at the bottom of page 5 in the block diagram. The compensation network sits at the output of the error amplifier.
The error amplifier outputs how far away the output voltage is from a target at any given time, which in this instance is 1.24V. At the target voltage your feedback divider would give 1.24V at the output. It uses this as a part of a calculation to adjust the duty cycle of the MOSFET to achieve the desired output.
Adding the RC filtering to this error signal helps keep the loop stable. If there was no compensation the loop would react so fast it would start to oscillate uncontrollably, as the output shot up and down the feedback system would keep overreacting to correct the output. This would be an underdamped condition.
If the compensation network slows down the error response too much on the other hand the regulator will be slow to react to changes at the output. For instance if something suddenly placed a heavy load on the regulator the output could sag very low before the regulator catches up. This would be an overdamped condition.
The goal of this network is to make sure the regulator can respond fast enough to react to the loads you will be placing on it, but not so fast enough it starts jerking itself around.
Linear and TI both have excellent app notes on current mode boost converter compensation which are a good place to start.
Here's one from linear: http://cds.linear.com/docs/en/application-note/AN149fa.pdf
From my personal experience many people use the given application circuit as a starting point, and then build out from there.
This document by TI is a great resource for understanding current mode control theory at a deeper level. Page 10 has an example of the control loop and transfer function. You can use that as a base and start add more circuit elements in.
Best Answer
If the datasheet doesn't explicitly give you the value of inductor to use, then go back to first principles and decide for yourself.
You say you want 6 V at 1 A out. You didn't specify the input voltage other than with the vague "Li-Po battery voltage", so I'll arbitrarily pick 3.5 V to use as example. From the datasheet, you can see that the IC switches at 1.6 MHz.
For efficiency, you want to run in continuous mode at the 1 A output current, and you want the ripple to be reasonably small relative to the average. However, if you make the ripple too small, then the controller will have a hard time regulating the output in response to transients. For sake of example, let's see what 100 mA peak to peak gets us.
When the switch is on, there will be just the 3.5 V input voltage applied to increase the current in the inductor. When the switch is off, the inductor is conducting current from the input to the output thru a Schottky diode. (6 V) - (3.5 V) = 2.5 V reverse voltage. However, the forward drop across the diode is also in there. Let's say that adds another 500 mV to the reverse voltage on the inductor, for a total of 3 V. So the inductor is being driven forwards with 3.5 V and reverse with 3.0 V.
Since the we are assuming 1 A average steady state, the current rise during the switch on time must equal the current drop during the switch off time. The switch on time will be (3.0 V)/(3.0 V + 3.5 V) = 46% of the total switching cycle, and the switch off time the remaining 54%.
The switching frequency is 1.6 MHz, so 625 ns period. From above, we find that the switch on time is 288 ns, and the off time 337 ns. Now we can simply plug in the numbers to get the inductor value that satisfies the constraints. Using the particulars from the switch on time:
H = V s / A = (3.5 V)(288 ns)/(100 mA) = 10.1 µH
As a sanity check, the particulars from the switch off time should yield the same answer:
(3.0 V)(337 ns)/(100 mA) = 10.1 µH
So, if you're OK with 100 mA peak to peak ripple at 1 A out, then get a 10 µH inductor.
You also have to decide the inductor current rating. In this example, the current will go from 950 mA to 1.05 A each cycle. That means the saturation current should be somewhat higher than 1.05 A, like maybe around 1.2 A at least.
You then get to trade off DC resistance with size and price. That's a judgement call you have to make. Lower DC resistance will yield higher efficiency. For example, the 10 µH part of the Coilcraft MSS6132 series has 70 mΩ resistance and a 10% drop saturation limit at 1.36 A. At 1 A, that's a extra 70 mV out of 3.0 to 3.5 V, or about 2% extra loss. Only you can decide whether that's good enough.