Write out the simple equations

$$\begin{align}
Y_j &= a_0 Z_j + a_1 Z_{j-1} + a_2 Z_{j-2}\\
Y_{j-1} &= a_0 Z_{j-1} + a_1 Z_{j-2} + a_2 Z_{j-3}
\end{align}$$

There are some very simple cases that make $Y_j \perp Y_{j-1}$ due to the independence assumption of the random variables $\{Z_i\}_{i\in\mathbb{Z}}$. An example is $a_0 \in \mathbb{R}\setminus \{0\},\, a_1 = 0,\, a_2 = 0$. Not sure if you were looking for a complete solution but this should help get you started. 

Also, an easy check for RV which are not independent is using the contrapositive form of the common theorem
$$X\perp Y \implies E[XY] = E[X]E[Y]$$
Note that the converse of this statement is not true. 

**Proof**

Assertion $a_1 = 0 \iff Y_j \perp Y_{j-1}$ 

($\implies$) Suppose not, so that $a_1 = 0$ and $Y_j$ is not independent of $Y_{j-1}$. 

$$\begin{align}
E[Y_j Y_{j-1}]  & = E[ a_0^2 Z_j Z_{j-1} + a_0 a_2Z_jZ_{j-3} + a_2 a_0 Z_{j-2} + a_2^2Z_{j-2}Z_{j-3}] \\
& = E[Y_j]E[Y_{j-1}]
\end{align}$$

This is a contradiction of the theorem above.

($\impliedby$) Suppose $Y_j \perp Y_{j-1}$ by theorem, we know that $E[Y_j Y_{j-1}] = E[Y_j]E[Y_{j-1}]$ calculating these values separately,

$$\begin{align}
E[Y_jY_{j-1}] = E[ &a_0^2 Z_j Z_{j-1} + a_0a_1 Z_j Z_{j-2} + a_0 a_2 Z_j Z_{j-3}\\
 & a_1 a_0 Z_{j-1}^2 + a_1^2 Z_{j-1}Z_{j-2} + a_1 a_2 Z_{j-1}Z_{j-3}\\
 & a_2 a_0 Z_{j-2}Z_{j-1} + a_2 a_1Z_{j-2}^2 + a_2^2Z_{j-2}Z_{j-3} ]
\end{align}$$

Notice that in the terms $a_1 a_0 Z_{j-1}^2$ and $a_2 a_1Z_{j-2}^2$ the RV is squared inside the expectation. Due to Jensen's inequality, the expectation of the squared value is not equal (in general) to the squared value of the expectation. Thus, $a_1 = 0$ for the theorem to hold.