Solution 1. Note that
$$\sin 3\theta = \sin \theta \cos 2\theta + \cos \theta \sin 2\theta = \sin \theta(1 - 2\sin^2 \theta) + \cos \theta (2\sin \theta \cos \theta) = 3 \sin \theta - 4 \sin^3 \theta.$$
So
$$3 \sin y - 4 \sin^3 y = - \sin x = -(3 \sin z - 4 \sin^3 z), \quad z = x/3.$$
Or with $t = 2 \sin y$, $s = 2 \sin z$,
$$3t - t^3 = -3s + s^3, \quad (t+s)(t^2 - ts + s^2 - 3) = 0.$$
Thus, either $t = -s$ or
$$t = \frac{s \pm \sqrt{s^2 - 4(s^2 - 3)}}{2} = \frac{s \pm \sqrt{3\sqrt{4-s^2}}}{2} = \sin z \pm \sqrt{3} |\cos z|,$$
Since
$$|\sin z \pm \sqrt{3} |\cos z|| \le \sqrt{1+3(\sin^2 z + \cos^2 z)} = 2,$$
this means that there are three possible values for $\sin y$, namely
$$-\sin\left(\frac{x}{3}\right), \quad \frac{1}{2}\left(\sin\left(\frac{x}{3}\right) \pm \sqrt{3}\cos\left(\frac{x}{3}\right)\right).$$

Solution 2. To solve the equation $\sin(3y) + \sin x = 0$, we use the trigonometric identity that converts a sum of sines to a product to get
$$2 \sin \left( \frac{3y+x}{2} \right) \cos \left( \frac{3y-x}{2} \right) = 0.$$
Therefore, $\sin((3y + x)/2) = 0$ or $\cos((3y - x)/2) = 0$, so
$$\frac{3y+x}{2} = n\pi \quad \text{or} \quad \frac{3y-x}{2} = \left(n+\frac{1}{2}\right)\pi \quad \text{for some } n \in \mathbb{Z}.$$
Hence $y = \frac{1}{3}(2n\pi - x)$ or $y = \frac{1}{3}((2n + 1)\pi + x)$, for some $n \in \mathbb{Z}$. So $\sin y = \sin(\frac{1}{3}(2n\pi - x))$ or $\sin y = \sin(\frac{1}{3}((2n+1)\pi+x))$, for some $n = 0, 1, 2$. Here we have used the periodicity of sine to restrict the values of $n$. Using the identity $\sin(\pi - \alpha) = \sin \alpha$, the latter three solutions can be written as $\sin y = \sin(\pi - \frac{1}{3}((2n+1)\pi - x)) = \sin(\frac{1}{3}((2 - 2n)\pi - x))$, for $n = 0, 1, 2$. These give the same as the first three solutions. So the three solutions are $\sin y = \sin(\frac{1}{3}(2n\pi - x))$, for $n = 0, 1, 2$, and this is the same as above.

$$-\sin\left(\frac{x}{3}\right), \quad \frac{1}{2}\left(\sin\left(\frac{x}{3}\right) \pm \sqrt{3}\cos\left(\frac{x}{3}\right)\right).$$