A special class of ODE that we study is of the form $y'=f(x,y)$ which is perhaps called differential equation of first order and first degree.Now often during differential equation course we are just given this definition and the professor mover towards discussing methods to tackle those problems mechanically and techniques of finding solution without mentioning what is actually going on.Here some questions may arise which are as follows:
1.What is the condition on $f$ is there any condition as $f$ must be continuous.
2.The most important one,what is the domain of the function $f$?
3.We are to find a function $y=\phi(x)$ satisfying the differential equation,but what is the domain on which we are finding $\phi$?
4.How do we know that those we find are precisely all of the solution and we missed none of them?
5.Consider differential equations like $xy'=y$,here we can divide by $x$ to get $y'=y/x$,why is division by $x$ allowed here,by dividing with $x$ we are implicitly assuming that $x$ cannot be $0$.This means we are excluding $0$ from the domain of $\phi$ and finding it on $x\neq 0$.Why are we doing that,would it not change the solution set?
If you want a theorem to tell you a solution exists locally, you'll want $f$ to be continuous. If you want a theorem to tell you the solution to an initial value problem is unique, you'll want it to be Lipschitz in $y$: continuously differentiable is sufficient.
But what you often have in practice is an $f$ that is good on most of the $xy$ plane, with some set (maybe a curve) where it is bad. Then you're OK (solutions exist and are unique) as long as your solution stays in the "good" set. It could be that the solution will always stay in the "good" set. Or it may be that the solution will hit the "bad" set at some point, and at that point it may cease to exist or become non-unique.
In your example $x y' = y$, the "bad" set is the $y$ axis. You can explicitly construct the solutions: $y = C x$ for constants $C$. These do satisfy the differential equation in the form $xy' = y$ for all $x$, even $x=0$, however since all of them go through the same point $(0,0)$ we see that uniqueness fails spectacularly at that point.