= Solution
Choose coordinates that measure the curved boundary and the slope of a ray:
$$
\boxed{u=xy^2,\qquad v=y/x.}
$$
On the positive interior, the inverse <change of variables> is
$$
x=u^{1/3}v^{-2/3},\qquad y=u^{1/3}v^{1/3}.
$$
The two rays become $v=a$ and $v=1$, while the curve becomes $u=1$. The interior of $S$ therefore maps to the rectangle $0<u<1$, $a<v<1$. For $a=0$, the lower edge $v=0$ describes a limiting boundary, since the original region is unbounded; the origin also has no uniquely defined ratio $y/x$. For $a=1$ the region has zero area. These endpoint cases do not invalidate the <change of variables> on the positive interior.
For the three-dimensional <integral>, retain $z$ as the third coordinate. The transformed region is
$$
0<z<1,\qquad z<v<1,\qquad0<u<1.
$$
The planar <Jacobian determinant> is
$$
\frac{\partial(u,v)}{\partial(x,y)}
=\det\begin{pmatrix}y^2&2xy\\-y/x^2&1/x\end{pmatrix}
=\frac{3y^2}{x},\qquad
\left|\frac{\partial(x,y)}{\partial(u,v)}\right|=\frac{x}{3y^2}.
$$
The integrand cancels this <Jacobian determinant> especially simply:
$$
\frac{y^2z^2}{x}\,dx\,dy\,dz=\frac{z^2}{3}\,du\,dv\,dz.
$$
Hence
$$
\boxed{\int_D\frac{y^2z^2}{x}\,dx\,dy\,dz
=\frac13\int_0^1z^2(1-z)\,dz=\frac1{36}.}
$$
One can first use the <change of variables formula> with $z\geq\varepsilon>0$ and then pass to the limit. The nonnegative integrand allows this by the <monotone convergence theorem>, so the unbounded limiting slice at $z=0$ causes no omitted contribution.
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