= Solution
First prove the general uniqueness claim. The bound $Mh\leq k$ prevents a solution from leaving the dependent-variable interval before $|x-a|=h$: up to any first exit, $|y(x)-b|\leq M|x-a|\leq k$. Let $y_1,y_2$ be two solutions with the same initial value. Their integral equations and the <Lipschitz condition> give, for $x\geq a$,
$$
|y_1(x)-y_2(x)|\leq A\int_a^x|y_1(s)-y_2(s)|\,ds.
$$
Since their difference is bounded by $2k$, repeated substitution into this inequality proves, for every integer $n\geq0$,
$$
|y_1(x)-y_2(x)|\leq2k\frac{(A|x-a|)^n}{n!}.
$$
The same argument with reversed limits works for $x\leq a$. 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 $f_y=(2/3)xy^{-1/3}$ is bounded, giving a uniform Lipschitz constant by the mean value theorem. For example $h=k=1/2$ works: $M\leq\tfrac12(3/2)^{2/3}$ and $Mh<1/2$. Separate variables using $u=y^{1/3}$, obtaining $3u'=x$. The initial condition gives $u=1+x^2/6$, hence
$$
\boxed{y(x)=\left(1+\frac{x^2}{6}\right)^3.}
$$
It remains positive for all real $x$, so the same local uniqueness argument continues along this solution.
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