Answer
$\dfrac{3}{4}$
Work Step by Step
We know
$$\lim\limits_{\theta \to 0} \dfrac{\mathrm{sin}\left(\theta \right)}{\theta }=1.$$
Thus for any nonzero real number $a$, it follows that
$$\underset{\theta \to 0}{\lim}\dfrac{\mathrm{sin}\left(a\theta \right)}{a\theta }=1.$$
We will use this fact to compute the limit below. So to see that this is true, we let $x=a\theta.$ Then since $x \to 0$ as $\theta \to 0$, we have $$\underset{\theta \to 0}{\lim}\dfrac{\mathrm{sin}\left(a\theta \right)}{a\theta }=\underset{x\to 0}{\lim}\dfrac{\mathrm{sin}\left(x\right)}{x}=1.$$
Now, we want to find $\underset{y\to 0}{\lim}\dfrac{\mathrm{sin}\left(3y\right)}{4y}$.
Note that
$$\dfrac{\mathrm{sin}\left(3y\right)}{4y}=\dfrac{3\mathrm{sin}\left(3y\right)}{3\left(4y\right)}=\dfrac{3}{4}\dfrac{\mathrm{sin}\left(3y\right)}{3y}.$$
Thus
$$\underset{y\to 0}{\lim}\dfrac{\mathrm{sin}\left(3y\right)}{4y}=\underset{y\to 0}{\lim}\left(\dfrac{3}{4}\frac{\mathrm{sin}\left(3y\right)}{3y}\right)=\dfrac{3}{4}\left(\underset{y\to 0}{\lim}\dfrac{\mathrm{sin}\left(3y\right)}{3y}\right)=\dfrac{3}{4}\left(1\right)=\dfrac{3}{4}.$$
