How to show: $y'(t)=y(t)\sin(\frac{1}{y(t)})$ has one well defined solution $\in C^1(\mathbb{R},\mathbb{R})$ We are supposed to show: $y'(t)=y(t)\sin(\frac{1}{y(t)})$ has one well defined solution $\in C^1(\mathbb{R},\mathbb{R})$ for some initial value $y(0)=y_{0}>0$.
There is also a hint: solve on a local interval with explicit bounds, then try to construct a global solution.
I have experimented with using Picard-Iteration to find solutions for explicit bounds, but I haven't seen a pattern. Obviously if $y_0=\frac{1}{n\pi}$ we get that a solution is the constant function $\varphi(x)=\frac{1}{n\pi}$. For other bounds, I can't even solve using iteration.
I don't see how I could use any other method. We can't use separation of variables as we only have one. It's not a linear differential equation.
What am I not seeing here?
 A: *

*So as indicated by Nikhil Sahoo, the Cauchy-Lipshitz (or Picard-Lindelof) theorem tells you that in order to prove the existence of an interval $[0,t_0]$ such that there exists a unique solution on it, you just have to prove that $f: z\mapsto z \sin(1/z)$ is locally Lipshitz.

*To prove that, remark that $f\in C^1(\mathbb{R}\backslash\{0\})$ as a product of composition of $C^1$ functions defined for $z\neq 0$. Then, to prove it in the neighbourhood of $0$, remark that $|f(z)|\leq |z|$ so that if $x\leq 0$ and $y>0$, you get
$$|f(x)-f(y)| \leq |x|+|y| = |x-y|.$$
Thus, this proves that $f$ is locally Lipshitz everywhere.

*Then to get global well-posedness, you just need to prove that your solution cannot blowup in finite time (according to a a theorem sometimes also known as a strong version of Picard-Lindelof theorem)
The shortest way to obtain this is to use again the fact that $|f(z)|\leq |z|$ so that for any $t$ where your solution is bounded $-y(t)\leq y'(t)\leq y(t)$, and by Gronwall's lemma, $|y(t)|\leq e^t$. Hence, your solution cannot blowup, which proves that your solution is global.
A: I am rewriting this answer, because there were some mistakes before and I've thought of some more stuff since. Firstly, let $f(y)=y\sin(1/y)$. This function is locally Lipschitz in $y$ and independent of $t$. Hence, the Picard-Lindelöf theorem gives local existence and uniqueness of solutions.
How do we prove the existence of a global solution? Well, we can extend the local solution over a maximal interval, which is some $(a,b)$ with $a<0<b$. The only reason that the solution would not extend to $a$ or $b$ is if it diverges out of the domain; in this case, that just means tending to $0$ or $\infty$ in finite time. So it remains to prove that this cannot happen. Before that, let's check out the solutions.
The following graphic is from Apple's "Grapher." As you noted, there are stationary points $1/\pi n$ for all positive integers $n$. These are the only stationary points, since they are precisely the zeros of $f$. When $n$ is odd, we get an unstable equilibrium point. When $n$ is even, we get a stable equilibrium point. If the initial value is $<1/\pi$, then the solution tends to a stationary point. But if the initial value is $>1/\pi,$ then the solution diverges to $\infty$ at an asymptotically linear rate. 
Now, how do we actually prove that the solutions extend? Firstly, we can't approach $0$, because we would have to pass a stationary point along the way (since there are arbitrarily small $1/\pi n$). Why don't we diverge to $\infty$ in finite time? If $x$ is very small, then $\sin (x)\approx x.$ Hence, if $y$ is very large, then $f(y)=y\sin(1/y)\approx y\cdot 1/y=1$. So once $y$ gets large enough, the system starts to look linear. Certainly then, we cannot have a vertical asymptote where $y$ approaches $\infty$ in finite time.

A: Can be solved in the same way as here. Let
$$f(y) = \int_{y_0}^y \frac {d\tau} {\tau \sin \frac 1 \tau}.$$
If $y_0 > 1/\pi$, then $f$ is strictly increasing from $-\infty$ to $\infty$ over $(1/\pi, \infty)$ because the integrals from $y_0$ to $1/\pi$ and from $y_0$ to $\infty$ diverge. Therefore $y = f^{-1}(t)$ is a solution with domain $\mathbb R$.
Any other $y_0$ except $0$ and $1/(\pi k)$ will also work.
