= Solution
Symmetry of the derivative matrix means $\partial_iF_j=\partial_jF_i$. Contracting this symmetric pair with the antisymmetric <Levi-Civita symbol> gives
$$
(\nabla\times\mathbf F)_k=\epsilon_{kij}\partial_iF_j=0.
$$
Every circle bounds its planar disk, and the <vector field> is smooth on all of $\mathbb R^3$, including that disk. <Stokes theorem> therefore gives \b[for either orientation]
$$
\boxed{\oint_C\mathbf F\cdot d\mathbf x=\int_D(\nabla\times\mathbf F)\cdot\mathbf n\,dS=0.}
$$
The global domain is essential to this argument; the same local symmetry condition on a domain with a hole need not make every closed-curve <line integral> vanish.
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