For example, the Apollo lunar module landing on the moon was worth approximately 60% fuel of the total weight (on wiki 8 tons of the 15 tons). Departure from the surface was about 50% of fuel (2.5 tons from 5 tons).
Assuming that you are talking about a propulsive Landing only, without parachutes, you would need around 3.8 kilometers per second of Delta V (from the map below). It should be noted that the amount of Delta V required to land on a body from orbit roughly equals the amount required to reach orbit. Source of image
To land: There is some air at Mars which will slow you down somewhat, so 3.8km/s of Delta V is a ballpark number which is higher than what it is in reality. As the drag due to air resistance will slow the vehicle down and therefore require less Delta V to land.
One major factor in knowing how much fuel you will actually need to perform the landing is the efficiency of the fuels. Taking into account the Apollo LEM which used Aerozine 50 and N2O4 and got an isp of 311 (3047m/s), it had around a 60% fuel mass to total mass ratio. But the moon requires much less delta V to land on than Mars. So a vehicle running the with the same fuels (N2O4 and Aerozine) would need a ratio of .72 fuel mass to total mass. However this ratio really depends on the efficiency of the engine, as a more efficient engine can have a lower ratio, but a less efficient engine will need a higher ratio. .72 which was calculated via the rocket equation does not take into account air resistance during landing (which will help in this case). So a lander would need at most a ratio of .72 for a propulsive landing only, neglecting air resistance.
To get to orbit from the surface: Since the amount of Delta V required to land from orbit is roughly the same to ascend from the surface to orbit, we will assume 3.8km/s of Delta V needed. It is helpful to not that this number is actually a lot closer to reality as air resistance will play much less of a role in ascent than descent. Let’s use the same fuels as the Apollo LEM once again and assume an efficiency of 311s (3047m/s).
The mass final in this case is around 2.8, giving a percentage of 72% (as mentioned earlier) for the fuel mass to total mass to reach Mars orbit. This is definitely lower than what would actually be used as this is the bare minimum to reach orbit, once in orbit the vehicle may need to rendezvous and use more fuel.
Once again, this ratio of .72 is only for the LEM descent engine. Use a more efficient engine and get lower ratios and vice versa.