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
$2Al^@ (s)$ + $3Zn^"2+"$ = $2Al^"3+"$ + $3Zn^@ (s)$
