"Methalox" requires a methane tank and a separate oxygen tank. "Hydrolox" requires a hydrogen tank and a separate tank of, again, oxygen. As liquified propellants absorb heat from their surroundings, their temperature will rise until they reach their respective boiling points. After that, the temperature of the liquid will stop rising but some of it will begin to phase transition into a gas.
If you created a propellant depo with three tanks containing liquid hydrogen, methane, and oxygen, and all three were at their boiling points, then the liquid hydrogen tank would be the coldest. You could take advantage of that and use the hydrogen tank to cool the other two tanks. Then you could keep the temperatures of the liquid methane and liquid oxygen tanks below their boiling points. This way you would only have to deal with boil-off in the liquid hydrogen tank - at least so long as there is still liquid hydrogen in that tank.
To figure out if this would be worthwhile, you need to know the enthalpy of vaporization of the three liquids. Hydrogen is 0.9 kJ/mol, Oxygen is 6.92 kJ/mol, and Methane is 8.17 kJ/mol.
If the liquid hydrogen absorbs 0.9 kJ, then one mol would be converted to H2 gas and so you'd lose about 2 grams.
If the liquid oxygen absorbs 0.9 kJ, then 0.9/6.92=0.13 mols of O2 would be converted to gas, about 4 grams.
If the liquid methane absorbs 0.9 kJ, then 0.9/8.17=0.11 mols of CH4 would be converted to gas, about 1.76 grams.
It might be worthwhile to send up some liquid H2 to keep the liquid O2 below its boiling point, but it probably would not be worthwhile to use liquid H2 to keep liquid CH4 below its boiling point.
If you did send up liquid H2 to cool the liquid O2, it might absorb heat faster since it's a lot colder than liquid oxygen. This would cancel out some of the benefits.
Still, it's an interesting idea...