How do I balance this equation using the "half reaction" method: #CH_4+O_2+H_2O=CO+CO_2+H_2# ?

Methane is oxidized to give #"(i) carbon monoxide"#, and #"(ii) carbon dioxide..........."#

#CH_4(g) + H_2O rarr CO(g) + 6H^(+) +6e^(-)#

#CH_4(g) +2H_2O rarr CO_2(g) +8H^(+) + 8e^(-)#

And overall, this gives..........

#2CH_4(g) +3H_2O rarr CO_2(g) +CO(g) +14H^(+) + 14e^(-)# #(i)#

But for oxidation there must be reduction.........and clearly dioxygen is the oxidizing agent (that is of course reduced..........to #O^(2-)#, a 2 electron reduction).

#O_2(g) +4e^(-) rarr 2O^(2-)# #(ii)#

And we balance this in the usual way; i.e. eliminate electrons, so #7xx(ii) +2xx (i)#, to give (finally):

#4CH_4(g) +7O_2(g) rarr 2CO_2(g) +2CO(g) +8H_2O(l)#

Which (I think) is balanced with respect to mass and charge, as indeed it must be if we purport to represent chemical, i.e. physical, reality.

Perhaps you should try this out for yourself, using the standard redox equation method that have been taught to you. The oxidation number of carbon in methane is #C^(-IV)#, of #CO# and #CO_2#, we have #C^(-+II)#, and #C^(+IV)#.

To assign oxidations numbers in higher alkanes, we recall that the definition of oxidation number, the charge left on the central atom, when all the bonds are broken, with the charge distributed to the most electronegative atom. For element-element bonds, the carbons have equal electronegativity, and thus for ethane, we have #C(-III)#, and for propane, we have #H_3stackrel(-III)C-stackrel(-II)CH_2-CH_3#, etc.