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)
