Determine whether the system is a FIR or IIR by looking the transfer function
The forward $z$-transform is written with negative powers on $z$. So, $z$ transform of your difference equation would be:

$$
Y(z) = \frac{1}{3}\left(zX(z) + X(z) + z^{-1}X(z)\right)
$$

yielding a transfer function of:

$$
H(z) = \frac{Y(z)}{X(z)} = \frac{1}{3}(z + 1 + z^{-1})
$$

The easiest way to determine whether a filter is IIR or FIR is to identify its pole locations. For FIR filters, there is a rule for this that is based on the structure of the impulse response:


If the system is causal (i.e. it is zero for all $n &lt; 0$), then it is FIR if all of its poles are located at the origin ($z=0$).
If the system is anticausal (i.e. it is zero for all $n \ge 0$), then it is FIR if all of its poles are located at $z=\infty$.


Your system is neither causal nor anticausal. Instead, it has a causal component ($\frac{1}{3}(x[n] + x[n-1])$) and an anticausal component ($\frac{1}{3}x[n+1]$). So, it can be helpful to look at each component separately:


$\frac{1}{3}(x[n] + x[n-1])$: The $z$-transform of the causal portion of the transfer function is $\frac{1}{3}(1 + z^{-1})$. This expression $\to \infty$ as $z \to 0$, so it has a pole at $z = 0$. This meets the criterion for an FIR response.
$\frac{1}{3}x[n+1]$: The $z$-transform of the anticausal portion of the transfer function is $z$. This expression $\to \infty$ as $z \to \infty$, so it has a pole at $z = \infty$. This meets the criterion for an FIR response. 


This would imply that you gave a difference equation for a non-causal FIR filter.