Since nobody posted a correct solution.

Observe first that the solution is unique since $f(t,y)= \frac{\sin{y}}{1+\sin^2(t)}$ satisfied conditions of (Theorem 2.4.2 in the textbook). Since $|f(t,y)\le 1$

solutions exists for $t\in (-\infty,+\infty)$.

Observe also that $f(t,y)=0$ iff $\sin(y)=0 \iff y=n\pi$ with $n=0,\pm 1, \pm 2,\ldots$ Therefore all other solutions cannot cross these values and remain confined between them; in particular, solution with $y(0)=1$ remains confined between $y=0$ and $y=2\pi$.

Since $y'=f(t,y)>0$ iff $2n\pi< y< (2n+1)\pi$ such solutions are monotone increasing. Since $y'=f(t,y)<0$ iff $(2n-1)\pi< y< 2n\pi$ such solutions are monotone decreasing. In particular, solution with $y(0)=1$ is monotone increasing.

Additional remarks: since $|f(t,y)|\ge \frac{1}{2}|\sin (\epsilon)|$ as $y \in (n\pi+\epsilon), (n+1)\pi-\epsilon)$ solutions will cross any other line and

* Any solution confined between $2n\pi $ and $(2n+1)\pi$, tends to $2n\pi$ as $t\to -\infty$ and to $(2n+1)\pi$ as $t\to +\infty$;

* Any solution confined between $(2n-1)\pi $ and $2n\pi$, tends to $2n\pi$ as $t\to -\infty$ and to $(2n-1)\pi$ as $t\to +\infty$.

Using language we learn in Chapter 9, $y=2n\pi$ are *asymptotically unstable stationary solutions*; $y=(2n+1)\pi$ are *asymptotically stable stationary solutions*. See attached picture