The equation given above is wrongly balanced, so I have to correct it first.
#Al^@# + #Zn(NO_3)_2# = #Al(NO_3)_3# + #Zn^@# (unbalanced)
Tallying the atoms based on subscripts,
left side: #Al# = 1; #Zn# = 1; #(NO_3)_3# = 2
right side: #Al# = 1; #Zn# = 1; #(NO_3)_3# = 3
Notice that I am considering the #NO_3^-# ion as one "atom" in order to avoid confusing myself.
Balancing the equations we have:
#color (blue) 2Al^@ (s) # + #color (red) 3 Zn(NO_3)_2# = #color (green) 2Al(NO_3)_3# + #color (magenta) 3Zn^@ (s)# (balanced)
left side: #Al# = (1 x #color (blue) 2#) = 2; #Zn# = (1 x #color (red) 3#) = 3; #(NO_3)_3# = (2 x #color (red) 3#) = 6
right side: #Al# = (1x #color (green) 2#) = 2; #Zn# = (1 x #color (magenta) 3#) = 3; #(NO_3)_3# = (3 x #color (green) 2#) = 6
Now that we have the correct balance equation, we must rewrite the equation showing all the ions involved in the reaction.
#2Al^@ (s) # + #3Zn^"2+"# + #6NO_3^-# = #2Al^"3+"# + #6NO_3^-# + #3Zn^@ (s)#
Notice that for the #NO_3^-#, I multiply the subscript with the substance's coefficient. Now let us get rid of your spectator ions (ions that are present in both sides of the equation).
#2Al^@ (s) # + #3Zn^"2+"# + #cancel (6NO_3^-)# = #2Al^"3+"# + #cancel (6NO_3^-)# + #3Zn^@ (s)#
Now let's show the electrons per half-reaction:
#2Al^@ (s) # = #2Al^"3+"# + #6e^-#
#3Zn^"2+"# + #6e^-# = #3Zn^@ (s)#
Thus, the net ionic equation is
#2Al^@ (s)# + #3Zn^"2+"# = #2Al^"3+"# + #3Zn^@ (s)#
