Asymptotic solution to $y'=\sin(xy)$ I need to prove that the following equation
$y'=\sin (xy)\tag{1}$
Has a solution $y\not\equiv 0$ such that $\lim\limits_{x\to+\infty}y=0$. I was able to conclude that any solution of this equation (except for $y\equiv 0$) cannot cross the line $y=0$, because by the theorem of existence and uniqueness for any $x_0\in\mathbb R$ there may only exist one solution satisfying starting condition $y(x_0)=0$, which is $y\equiv 0$. This means that, for example, if a solution has a point with a value greater than zero, then the entire solution is greater than zero and it is bounded from below.
However, I am not sure where to go next. Even if I could prove that a solution is monotonically decreasing starting at some point, being bounded by zero doesn't guarantee that the limit equals zero. Any help would be appreciated
 A: Whenever $x$ is sufficiently large and $xy$ is close to $(2n+1)\pi$, the equation becomes approximately
$$
y'(x)=(-1)^{2n+1}\sin(xy-(2n+1)\pi)\approx(2n+1)\pi-xy,
$$
using $\sin(u+k\pi)=(-1)^k\sin(u)$ and the small-angle approximation for $xy-(2n+1)\pi\approx 0$.
The approximate equation has a solution via integrating factor $e^{x^2/2}$
$$
y(x)=Ce^{-x^2/2}+(2n+1)\pi e^{-x^2/2}\int_0^xe^{s^2/2}ds=Ce^{-x^2/2}+(2n+1)\pi\sqrt{2}D(x/\sqrt2)
$$
where $D$ is the Dawson function, a reduction of the imaginary error function. As can be found at the link, for large $x$ one has asymptotically $D(x)\sim\frac1{2x}$. As the first term vanishes quickly, this results in $y(x)\sim\frac{(2n+1)\pi}{x}$, so that the solution remains close to equilibrium of the right side, and thus also inside the assumptions of the initial approximation of the DE.
A suitable numerical visualization confirms this result, the plot below contains the plots of $xy(x)/\pi$ for several initial values $y(10)$. It is well visible that there are attracting lines close to the odd integers while the even integers are unstable.

