Let #a,b,c,d,e,f# be the number of molecules of each reactant. Th echecmical eqaution becomes

#aCuS+bHNO_3->cCu(NO_3)_2+dS+eH_2O+fNO#

Obsrvation reveals that nmber of molecules of #H NO_3# must be an even number to accommodate #H_2# on the right of the equation.

Let's start balancing for the radical #NO_3#.

On the right hand side of the equation the radical is either part of

atom #Cu(NO_3)_2# or can be obtained from combination of #H_2O# and #NO#. We may choose lowest integer 1 for either and double the number of atoms of the other. However, choosing lowest integer 1 for #NO# will rquire chosing 2 atoms of #H_2O#. This will increase the number of atoms of #H# required for balancing. Therefore, choosing 2 atoms of #NO# or #f=2#, the equation becomes

#aCuS+bHNO_3->cCu(NO_3)_2+dS+eH_2O+2NO#

With this we need 3 more atoms of #O# for the second atom of #N#. Implies #e=4#. The equation becomes

#aCuS+bHNO_3->cCu(NO_3)_2+dS+4H_2O+2NO#

Now after having fixed number of atoms of #H# on the right side, it follows that #b # must#=8#. To give us

#aCuS+8HNO_3->cCu(NO_3)_2+dS+4H_2O+2NO#

Now balacaing 24 number of #O# atoms of left side with the right side atoms we obtain #c=3#. This fixes #a=3=d#. We obtain the balanced equation as

#3CuS+8HNO_3->3Cu(NO_3)_2+3S+4H_2O+2NO#