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In Cartesian components, $\mathbf{F}=(z-y,x,-x)$, soIt cannot be a conservative vector field. For the second field, put $\Phi (\mathbf{r})=-(1+|\mathbf{r}{|}^2{)}^{-1/2}$. Its gradient is $\nabla \Phi =\mathbf{r}/(1+|\mathbf{r}{|}^2{)}^{3/2}=\mathbf{G}$. This smooth global scalar function proves $\mathbf{G}$ is conservative, with $\boxed{\nabla \times \mathbf{G}=0}$. The following path integrals exhibit the different behavior directly.

For two independent smooth constraints $g=h=0$, the tangent direction is orthogonal to both gradients. At a constrained stationary point, the gradient of the objective is also orthogonal to that direction, hence lies in their span: $\nabla f=\lambda \nabla g+\mu \nabla h$. This is the two-constraint Lagrange multiplier condition. Alternatively, distance extrema on a central plane section of an ellipsoid follow from a two-dimensional eigenvalue problem.

Minimizing distance is equivalent to minimizing $f=x^2+y^2+z^2$. Put $g=x-y+z$ and $h=x^2+(y^2+z^2)/4-1$. The Lagrange multiplier equations areAdding the latter two gives $(2-\mu /2)(y+z)=0$. If $\mu =4$, those equations give $\lambda =x=0$; the constraints then give $y=z=\pm \sqrt{2}$ and $f=4$. Otherwise $z=-y$ and $x=2y$. The ellipsoidal constraint gives $y=\pm \sqrt{2}/3$, and henceThe intersection is compact. The two constraint gradients are independent there: if $\nabla h=t\nabla g$, then $(x,y,z)=(t/2,-2t,2t)$, and $g=0$ forces $t=0$, contradicting $h=0$. Thus every global extremum is among these candidates. Taking square roots gives

For $k>0$, the two ordinary differential equations have solutionsTheir products $h(x)g(y)$ and $g(x)h(y)$ solve Laplace's equation, since the two second derivatives contribute opposite $k^2$ terms. For $k=0$, both factors are linear functions. This is separation of variables with positive, negative and zero separation constants.