= Solution
On the part with $1\leq z\leq2$, write
$$
\mathbf r(\rho,\phi)
=\left(\rho\cos\phi,\rho\sin\phi,
\sqrt{\rho^2-1}\right),
\qquad
\sqrt2\leq\rho\leq\sqrt5.
$$
This is the upper portion of a <one-sheet hyperboloid>[hyperboloid of one sheet]. Choosing the <orientation of a surface>[outward orientation], whose normal points radially away from the $z$-axis, the <vector area element> is
$$
\begin{aligned}
d\mathbf S
&=(\mathbf r_\phi\times\mathbf r_\rho)\,d\rho\,d\phi\\
&=\boxed{\left(
\frac{\rho^2\cos\phi}{\sqrt{\rho^2-1}},
\frac{\rho^2\sin\phi}{\sqrt{\rho^2-1}},
-\rho\right)d\rho\,d\phi}.
\end{aligned}
$$
Its radial component is positive and its vertical component is negative. Reversing the normal reverses all the fluxes below.
For
$$
\mathbf A=(-yz^2,xz^2,0),
$$
the <curl> is
$$
\nabla\times\mathbf A=(-2xz,-2yz,2z^2).
$$
On the surface, $z^2=\rho^2-1$, and therefore
$$
(\nabla\times\mathbf A)\cdot d\mathbf S
=-2\rho(2\rho^2-1)\,d\rho\,d\phi.
$$
Direct integration gives
$$
\begin{aligned}
\int_S\nabla\times\mathbf A\cdot d\mathbf S
&=-2\int_0^{2\pi}\int_{\sqrt2}^{\sqrt5}
\rho(2\rho^2-1)\,d\rho\,d\phi\\
&=-2\pi[\rho^4-\rho^2]_{\sqrt2}^{\sqrt5}\\
&=\boxed{-36\pi}.
\end{aligned}
$$
To verify this with the <Stokes theorem>, note that on a circle of fixed $\rho,z$,
$$
\mathbf A=z^2\rho\,\mathbf e_\phi,
\qquad
d\mathbf r=\rho\mathbf e_\phi\,d\phi.
$$
The induced <boundary orientation> runs in the negative $\phi$ direction on the top circle and the positive $\phi$ direction on the bottom circle. Hence
$$
\oint_{\partial S}\mathbf A\cdot d\mathbf r
=2\pi(1^2)(\sqrt2)^2
-2\pi(2^2)(\sqrt5)^2
=4\pi-40\pi
=-36\pi.
$$
For $S'$, Stokes' theorem again reduces the flux to its two boundary circles. The outward orientation gives positive $\phi$ direction at $z=-1$, where $\rho^2=2$, and negative $\phi$ direction at $z=\sqrt2$, where $\rho^2=3$. Thus
$$
\boxed{
\int_{S'}\nabla\times\mathbf A\cdot d\mathbf S
=2\pi(1)(2)-2\pi(2)(3)
=-8\pi}.
$$
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