= Solution
Away from the vertex, use the defining <function> $H=z^2-x^2-y^2$. Its <gradient> is normal to the smooth part of the <double circular cone>, and
$$
\mathbf F\cdot\nabla H=(x,y,z)\cdot(-2x,-2y,2z)=2H=0
$$
on the surface. Hence \b[the field is tangent at every regular point of the cone]. At the origin the cone has no unique tangent plane, but $\mathbf F(0)=0$, which belongs to its tangent cone. An equivalent argument, valid also at the vertex, uses the <integral curves of a vector field>: $\dot{\mathbf x}=\mathbf x$ has solution $\mathbf x(t)=e^t\mathbf x(0)$, and these dilations preserve $H=0$. This is an example of <radial dilation tangency to homogeneous zero sets>.
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