= Solution
Use the compatible open semicircles and set $r=\sqrt{x^2+y^2}$. The <characteristic curve> through $(\cos\theta,\sin\theta,-1)$ has
$$
z(s)=-(1-s)^{-1/3},\qquad s<1.
$$
The condition $s<1$ is essential: at $s=1$ there is <finite-time blowup>, so the opposite branch cannot be reached from the data. From $r=(1-s\sin\theta)^{-1}$ and $y=r\sin\theta$,
$$
s=\frac{r-1}{y}.
$$
Eliminating $s$ yields
$$
\boxed{z(x,y)=-\left(\frac{y}{1+y-\sqrt{x^2+y^2}}\right)^{1/3}.}
$$
The ratio inside the <cube root> is positive precisely on the two data-determined regions
$$
\boxed{\begin{aligned}
D_+&=\{(x,y):y>0,\ r<1+y\}
=\{(x,y):y>0,\ y>(x^2-1)/2\},\\
D_-&=\{(x,y):y<0,\ r>1+y\}
=\{(x,y):y<0,\ y<(x^2-1)/2\}.
\end{aligned}}
$$
On $D_+$ each point traces to the upper semicircle, and on $D_-$ to the lower semicircle, with $s<1$. To check the parameter ranges explicitly, for $\eta=\sin\theta>0$ they are $-\infty<s<1$, whereas for $\eta<0$ they are $1/\eta<s<1$. At the latter lower endpoint the planar projection escapes to infinity. The <inverse function theorem> and uniqueness for the characteristic <ordinary differential equations> give a unique smooth solution on each region. Equivalently, each point has the unique foot $(x/r,y/r)$ and parameter $(r-1)/y$.
A direct check is particularly short. For $s=(r-1)/y$, let $\mathcal D=xy\partial_x+y^2\partial_y$. Then $\mathcal Dr=yr$, $\mathcal Dy=y^2$, and
$$
\mathcal Ds=\frac{(yr)y-(r-1)y^2}{y^2}=1.
$$
Since $dz/ds=-z^4/3$, the displayed formula solves the <partial differential equation>, and $r=1$ gives $z=-1$ on both open semicircles. The finite boundary where $r=1+y$, away from the axis, is the <parabola> $y=(x^2-1)/2$ and corresponds to blowup. Points on $y=0$ cannot be added to a classical continuation of these regions: where they are approached from a data-determined region, the height behaves like a nonzero multiple of $|y|^{1/3}$, or has incompatible limits at the endpoints. The formula is therefore the complete classical answer for the compatible arcs, while the full-circle problem as printed has no classical solution.
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