Direct form I and II, y(n)=−0.1y(n−1)+0.2y(n−2)+3x(n)…

The printed equation has x(n−2) twice. I read it as y(n) = −0.1y(n−1) + 0.2y(n−2) + 3x(n) + 3.6x(n−1) + 0.6x(n−2) (the second term should be x(n−1)). If it is taken literally, the two x(n−2) terms combine to 4.2x(n−2) and only b₁ = 0, b₂ = 4.2 change.

H(z) = (3 + 3.6z⁻¹ + 0.6z⁻²) / (1 + 0.1z⁻¹ − 0.2z⁻²)

Direct Form I

Separate delays for input and output (2 + 2 = 4 delays).

 x[n]   ─────►[×3]──────────┐
   │z⁻¹                     │
 x[n-1] ─────►[×3.6]────────┤
   │z⁻¹                     ▼
 x[n-2] ─────►[×0.6]─────►( Σ )────┬──► y[n]
                            ▲      │
 y[n-1] ─────►[×−0.1]───────┤      │
   │z⁻¹    (y[n] is delayed to      │
 y[n-2] ─────►[×0.2]────────┘      │ give y[n-1], y[n-2])

In words: y[n] = 3x[n] + 3.6x[n−1] + 0.6x[n−2] + (−0.1)y[n−1] + (0.2)y[n−2], with x[n−1], x[n−2] from a two-delay line on the input and y[n−1], y[n−2] from a two-delay line on the output.

Direct Form II

Let w[n] = x[n] − 0.1w[n−1] + 0.2w[n−2], and y[n] = 3w[n] + 3.6w[n−1] + 0.6w[n−2].

 x[n]──►(+)────────┬──────►[×3]────►(+)──┬──► y[n]
         ▲   w[n]  ▼ z⁻¹                 ▲  │
         │         ├────────►[×3.6]──────┤
         ├─[×−0.1]─┤                     │
         │         ▼ z⁻¹                 │
         │         └────────►[×0.6]──────┘
         └─[×0.2]──┘

(−0.1 and 0.2 are the feedback multipliers on w[n−1] and w[n−2]; 3, 3.6, 0.6 are the feed-forward multipliers.)

Direct Form II needs only 2 delays (canonic), against 4 for Direct Form I.

Answer: DF-I: 4 delays; DF-II: 2 delays; feedback −0.1, 0.2; feed-forward 3, 3.6, 0.6.