If you try to plot this graph:
$$\sin x+\sin y=\frac{y^2}x$$
You will get an odd-looking shape:
But if you zoom out a little you can see a kind of cycle with squashed "squares":
But one of them has a "drop" "falling" from it, it seems to start around $\pm13.053$
If we zoom out a little more we discover a cycle of drops and furthermore, we see a new cycle of drops starts, this starts around $\pm32.334$
Where exactly do those "drops" create and why?
First, let's multiply the equation by $x$ to get rid of the singularity when $x=0$. So, we want to study $$x(\sin x+\sin y)=y^2.$$
A way to try to understand this is that we have two surfaces in 3D ($z=x(\sin x+\sin y)$ and $z=y^2$) and we want to know where they intersect. Let's take a look:
The yellow surface is $z=x(\sin x+\sin y)$ and the blue surface is $z=y^2$. The first surface is sort of like an infinite egg carton, and where the second surface flattens out the peaks tend to intersect it in closed curves that are sort of rounded squares (the "drops"). As $x$ gets bigger, the peaks of the first surface get taller and taller and so more and more will intersect the second surface.
We can predict where the "drops" occur in the following way. Since the value of the sine function lies in the interval $[-1,1]$, we know $$-2\lvert x\rvert\leq x(\sin x+\sin y)\leq 2\lvert x\rvert$$ for all $x,y$. Thus, wherever the two surface intersect, we know that $-2\lvert x\rvert\leq y^2\leq 2\lvert x\rvert$ will hold (though not conversely of course!), and since $y^2\geq 0$ we can simplify this to $y^2\leq 2\lvert x\rvert$.
Thus, all the "drops" lie inside the region given by $\lvert x\rvert\geq \tfrac{1}{2}y^2$. Indeed, here is a plot showing the they lie in the region bounded by the orange curve:
To predict where the drops occur, we could look at where the maxima of $x(\sin x+\sin y)$ are using some calculus, and then it should be that if the maximum lies inside the above region, that is the start of a new "cycle of drops."