Yes, that's correct. Another way to see $\operatorname{curl}F = 0$ is to realize that it's the gradient of the scalar field $\phi(x,y,z) = xy + yz + zx$.
If you want to use Stokes,
$$
\int_C F\cdot dr=\int\!\!\!\int_S\text{curl} F\cdot dS,
$$
where $S$ is a surface that has $C$ as boundary.
The advantage here is that
$$
\text{curl}\,F=\begin{vmatrix}i&j&k\\ D_x& D_y& D_z\\-y^2&x&z^2\end{vmatrix}
=(0,0,1+2y)
$$
is simple.
If we take $S$ as the disc in the plane $y+z=2$ and bounded by the cylinder, we can parametrize it by
$$
u(r,t)=(r\cos t, r\sin t, 2-r\sin t),\ \ 0\leq r\leq1,\ 0\leq t\leq 2\pi.
$$
Then normal vector is
$$
u_r\times u_t=\begin{vmatrix}i&j&k\\ \cos t& \sin t& -\sin t\\ -r \sin t& r\cos t&-r\cos t\end{vmatrix}
=(0,r,r).
$$
Then
$$
\int_C F\cdot dr=\int\!\!\!\int_S\text{curl} F\cdot dS
=\int_0^1\int_0^{2\pi}(0,0,1+2r\sin t)\cdot(0,r,r)\,dt\,dr\\
=\int_0^1\int_0^{2\pi}(r+2r^2\sin t)\,dt\,dr
=\int_0^12\pi r\,dr=\pi.
$$
In this case, it is also easy to solve the line integral directly: using the parametrization $c(t)=(\cos t, \sin t,2-\sin t)$, $0\leq t\leq 2\pi$, we get
$$
\int_C F\cdot dr=\int_0^{2\pi}(-\sin^2t,\cos t,(2-\sin t)^2)\cdot(-\sin t,\cos t,-\cos t)\,dt=\int_0^{2\pi}(-\sin^3t+\cos^2 t-(2-\sin t)^2\cos t )\,dt=\pi.
$$
Best Answer
Please note in this specific case, your vector field is gradient of scalar field $F$
where $F = \frac{x^2y^2}{2} + \frac{z^4}{4}$.
$(xy^2, x^2y, z^3) = \nabla(\frac{x^2y^2}{2} + \frac{z^4}{4})$
As it is in conservative vector field, the line integral over a closed loop will be zero.
But if it was not, please note that you are interested in intersection curve of $x^2 + y^2 = 4$ and $x + y + z = 1$. The unit normal vector to the plane will be $(\frac{1}{\sqrt3}, \frac{1}{\sqrt3}, \frac{1}{\sqrt3}).$
You then need to find curl of the vector field $\nabla \times F$ normal to the surface (do a dot product with unit normal) and then do the double integral.