= Solution
The <directional derivative> along the specified <vector field> gives
$$
(\mathbf F\cdot\nabla)\mathbf F=(x,y,z),\qquad
\mathbf F\times(\mathbf F\cdot\nabla)\mathbf F=(2yz,-2xz,0).
$$
Use the magnitude formula for the <curvature of an integral curve of a vector field>:
$$
\boxed{\kappa(x,y,z)=\frac{2|z|\sqrt{x^2+y^2}}{(x^2+y^2+z^2)^{3/2}},\qquad (x,y,z)\ne(0,0,0).}
$$
It vanishes on the <plane> $z=0$ and on the $z$ axis away from the origin, consistent with straight field lines there. At the origin the <vector field> vanishes, so it fails the nowhere-zero hypothesis and this field-line curvature is not defined.
As an independent geometric check, an <integral curve of a vector field> satisfies $\dot x=x$, $\dot y=y$, $\dot z=-z$, giving $(x,y,z)=(ae^u,be^u,ce^{-u})$. Its velocity is $(x,y,-z)$ and acceleration $(x,y,z)$; the ordinary <curvature of a space curve> formula $|r'\times r''|/|r'|^3$ gives the same expression.
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