Answer
The first $N$ has an oxidation number equal to $(-3)$
Each $H$ has that equal to $(+1)$
The other nitrogen has an oxidation number equal to $(+5)$
Each $O$ has an oxidation number equal to $(-2)$
Work Step by Step
We can separate $NH_4NO_3$ in 2 ions:
$NH{_4}^+$ and $N{O_3}^-$.
1. $N{H_4}^+$.
According to rule 6(d), the oxidation number for hydrogen is equal to $(+1)$
Using rule 4, and calling the oxidation number for $N$ "x":
$(x) * 1 + (+1)*4 = +1$
x + 4 = + 1
x = 1 - 4
x = -3
The oxidation number for the nitrogen in $N{H_4}^+$ is $(-3)$
2. $N{O_3}^-$.
According to Rule 7(a), each oxygen has an oxidation number of $(-2)$.
Using rule 4 and calling the oxidation number for $N$ "y":
$(y) * 1 + (-2) * 3 = -1$
y - 6 = -1
y = + 5
