= Solution
The first field is
$$
B_1=\nabla\bigl(e^{x^2}\cos(yz)\bigr),
$$
so its <integral> around $C$ is zero and it is conservative with potential $e^{x^2}\cos(yz)$.
Along $C$, $dr=(-\sin s,\cos s,0)ds$ and
$$
B_2=(-\sin s,\cos s,0),
$$
so
$$
\oint_CB_2\cdot dr=\int_0^{2\pi}1\,ds=\boxed{2\pi}.
$$
Thus $B_2$ is not conservative and cannot be a <gradient> on the stated punctured region.
Finally,
$$
B_3=\nabla\log(x^2+y^2+z^2).
$$
Its <integral> around $C$ is zero, and it is conservative with the displayed potential. Therefore the three <integrals> are respectively
$$
\boxed{0,\quad2\pi,\quad0},
$$
and exactly $B_1,B_3$ are conservative <gradient> fields.
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