Consider $y'=3y^{2/3},y(0)=0$ and let $R$ be the rectangle $|x|\leq1,|y|\leq1.$ Here $f(x,y)=3y^{2/3}$ is plainly continuous on $R$.
Also $y_1(x)=x^3$ and $y_2(x)=0$ are two different solutions valid for all $x$ thus the equation is not unique.
The non uniqueness lies in the fact that $f(x,y)$ does not satisfy a Lipschitz condition on the rectangle $R$, since the difference quotient
$\frac{f(0,y)-f(0,0)}{y=0}=\frac{3y^{2/3}}{y}=\frac{3}{y^{1/3}}$ (*)
is unbounded in every neighborhood of the origin.
This is the explanation in my textbook but I have some questions about it and would appreciate some clarification.
1) How did they obtain the two solutions, $y_1(x), y_2(x)$?
2) How do you know which values to plug into equation (*) and how does this help you see uniqueness?
$y'=3y^{2/3},y(0)=0$ can be solved separating variables
$\dfrac{dy}{dx}=3y^{2/3}\rightarrow \dfrac{dy}{3y^{2/3}}=dx$
Integrating both sides $y^\frac13=x+c\rightarrow y=x^3$
$c=0$ for the initial condition
The solution $y=0$ is obvious: $0=3\cdot 0^\frac23$