Start by assigning oxidation numbers to all the atoms that take part in the reaction
#stackrel(color(blue)(+1))(H_2) stackrel(color(blue)(-2))(S_((s))) + stackrel(color(blue)(+5))(N) stackrel(color(blue)(-2))(O_(3(aq))^(-)) -> stackrel(color(blue)(0))(S_(8(s))) + stackrel(color(blue)(+2))(N) stackrel(color(blue)(-2))(O_((g)))#
Notice that sulfur goes from an oxidation state of -2 on the reactants' side, to an oxidation state of zero on the products' side #-># sulfur is being oxidized.
Likewise, nitrogen's oxidation state changes from +5 to +2 #-># nitrogen is being reduced.
The oxidation and reduction half-reactions will look like this
#H_2 stackrel(color(blue)(-2))(S) -> stackrel(color(blue)(0))(S_8) + 2e^(-)#
Balance the sulfur atoms by multiplying the left side of the reaction by 8
#8H_2 stackrel(color(blue)(-2))(S) -> stackrel(color(blue)(0))(S_8) + 16e^(-)#
Balance the hydrogen atoms by adding 16 protons, #H^(+)#, to the products' side
#8H_2 stackrel(color(blue)(-2))(S) -> stackrel(color(blue)(0))(S_8) + 16e^(-) + 16H^(+)#
#stackrel(color(blue)(+5))(N) O_3^(-) + 3e^(-) -> stackrel(color(blue)(+2))(N) O#
Balance the oxygens by adding 2 water molecules to the products' side
#stackrel(color(blue)(+5))(N) O_3^(-) + 3e^(-) -> stackrel(color(blue)(+2))(N) O + 2H_2O#
Now balance the hydrogens by adding 4 protons to the reactants' side
#4H^(+) + stackrel(color(blue)(+5))(N) O_3^(-) + 3e^(-) -> stackrel(color(blue)(+2))(N) O + 2H_2O#
In a redox reaction, the number of electrons lost during oxidation and gained during reduction must be equal. At this point, you have #3e^(-)# gained in the reduction half-reaction, and #16e^(-)# lost during the oxidation half-reaction.
Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 16 to balance the electrons out
#{ (24H_2 stackrel(color(blue)(-2))(S) -> 3stackrel(color(blue)(0))(S_8) + 48e^(-) + 48H^(+)), (62H^(+) + 16stackrel(color(blue)(+5))(N) O_3^(-) + 48e^(-) -> 16stackrel(color(blue)(+2))(N) O + 32H_2O) :}#
Add the two half-reactions and cancel the species present on both sides of the equation to get
#24H_2S + 16H^(+) + cancel(48e^(-)) + 16NO_3^(-) ->3S_8 + cancel(48e^(-)) + 16NO + 32H_2O#
The balanced equation will thus be
#24H_2S + 16H^(+) + 16NO_3^(-) -> 3S_8 + 16NO + 32H_2O#
