 Solution

First prove the general uniqueness claim. The bound $Mh\le k$ prevents a solution from leaving the dependent-variable interval before $|x-a|=h$: up to any first exit, $|y(x)-b|\le M|x-a|\le k$. Let $y_1,y_2$ be two solutions with the same initial value. Their integral equations and the Lipschitz condition give, for $x\ge a$,Since their difference is bounded by $2k$, repeated substitution into this inequality proves, for every integer $n\ge 0$,The same argument with reversed limits works for $x\le a$. Letting $n\to \infty$ forces the difference to zero on the whole interval. This proves at most one solution without assuming that $Ah<1$.

For case (i), choose a rectangle around $(0,1)$ that stays in $y>0$. There $f_y=(2/3)x{y}^{-1/3}$ is bounded, giving a uniform Lipschitz constant by the mean value theorem. For example $h=k=1/2$ works: $M\le \frac{1}{2}(3/2{)}^{2/3}$ and $Mh<1/2$. Separate variables using $u=y^{1/3}$, obtaining $3u^{\prime }=x$. The initial condition gives $u=1+x^2/6$, henceIt remains positive for all real $x$, so the same local uniqueness argument continues along this solution.