= Solution
<Continuity> on the <compact> rectangle gives finite constants $M=\max_R|F|$ and $L=\max_R|F_y|$. The real <mean value theorem> in the second coordinate gives $|F(x,y)-F(x,z)|\le L|y-z|$ within the rectangle. Define the <Picard iterate>[Picard iterates] on the entire interval by
$$
y_0(x)=b,\qquad y_{n+1}(x)=b+\int_a^xF(t,y_n(t))dt.
$$
If the previous iterate stays in the vertical range of $R$, then $|y_{n+1}(x)-b|\le M|x-a|\le Mh\le k$. Induction therefore makes all these iterates well defined on $|x-a|\le h$.
Repeated use of the <Lipschitz bound> gives the factorial estimate
$$
|y_{n+1}(x)-y_n(x)|\le\frac{ML^n|x-a|^{n+1}}{(n+1)!}\quad(n\ge0).
$$
The case $n=0$ is the direct integral bound; integrating the estimate for $n-1$ proves the next one, on either side of $a$. The series of uniform bounds converges, so the iterates converge uniformly to a <continuous function> $y$ taking values within the closed vertical range. The Lipschitz estimate permits passage through the integral to give $y(x)=b+\int_a^xF(t,y(t))dt$. The <fundamental theorem of calculus> now gives $y'=F(x,y)$ and $y(a)=b$. Thus the solution exists on the full stated interval; no extra restriction $Lh<1$ was used.
For the specified equation take $k=1/4$, so $3/4\le y\le5/4$. On $|x|\le h$,
$$
M=\frac{3h^2}{4(3/4)^3}=\frac{16h^2}{9}.
$$
The condition $Mh\le k$ holds at $h^3=9/64$, namely $\boxed{h=3^{2/3}/4}$. The rectangle avoids $y=0$, so both <continuity> hypotheses hold there.
Direct separation gives $4y^3y'=3x^2$, hence $(y^4)'=(x^3)'$ and the initial value yields $y^4=1+x^3$. The branch through $y(0)=1$ is
$$
\boxed{y(x)=(1+x^3)^{1/4}\quad(x>-1).}
$$
It remains positive on this interval. At $x=-1$ it reaches zero, where the differential equation is undefined, so this is its maximal real interval containing zero. The rectangle theorem supplies a sufficient interval, smaller than that actual interval.
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