Evaluate \( \lim _{\theta \rightarrow 0} \frac{\sin ^{2}(3 \theta)}{\theta^{2}} \). If the corresponding limit is infinity, then insert INF in the blank. If the corresponding limit is negative infinity, then insert -INF in the blank. If the corresponding limit does not exist, then insert DNE in the blank. Otherwise, write the number as an integer or a reduced fraction.

To evaluate the limit \( \lim _{\theta \rightarrow 0} \frac{\sin ^{2}(3 \theta)}{\theta^{2}} \), we start by rewriting the expression in a more manageable form. Using the fact that \( \sin(x) \) can be approximated by \( x \) as \( x \rightarrow 0 \), we have: \[ \sin(3\theta) \approx 3\theta \text{ when } \theta \rightarrow 0. \] Thus, we can write: \[ \sin^2(3\theta) \approx (3\theta)^2 = 9\theta^2. \] Now substituting this approximation into the limit gives: \[ \lim_{\theta \rightarrow 0} \frac{\sin^2(3\theta)}{\theta^2} \approx \lim_{\theta \rightarrow 0} \frac{9\theta^2}{\theta^2} = \lim_{\theta \rightarrow 0} 9 = 9. \] Thus, the limit evaluates to: \[ \lim _{\theta \rightarrow 0} \frac{\sin ^{2}(3 \theta)}{\theta^{2}} = 9. \] Therefore, the answer is: \[ \boxed{9}. \]