The ratio of coefficients of $HN{O}_3,Fe{\left(N{O}_3\right)}_2$  and $N{H}_{4}N{O}_3$  in the  redox equation $Fe+HN{O}_3\to Fe{\left(N{O}_3\right)}_2+N{H}_{4}N{O}_3+H_{2}O$ in the balanced from will be

$\stackrel{0}{Fe}+\stackrel{+5}{HN{O}_3}\to \stackrel{+2}{Fe}{\left(N{O}_3\right)}_2+\stackrel{-3}{N{H}_4}N{O}_2+H_{2}O$

$\left(i\right)\stackrel{0}{Fe}\to \stackrel{+2}{Fe}{\left(N{O}_3\right)}_2\downarrow 2$

$\left(ii\right)\stackrel{+5}{HN{O}_3}\to \stackrel{-3}{N{H}_4}N{O}_3\downarrow 8$

To make increase in O.N.=decrease, multiply eqn.

(i) by 4 and add. Hence

$4Fe+HN{O}_3\to 4Fe{\left(N{O}_3\right)}_2+N{H}_{4}N{O}_3$

To balance N atoms on the two sides, multiply $HN{O}_3$  by 10. Hence

$4Fe+10HN{O}_3\to 4Fe{\left(N{O}_3\right)}_2+N{H}_{4}N{O}_3$

To balance O-atoms, add $3H_{2}O$  on RHS. Hence

$4Fe+10HN{O}_3\to$

                          $4Fe{\left(N{O}_3\right)}_2+N{H}_{4}N{O}_3+3H_{2}O$

H atoms are automatically balanced.