 Solution

A sufficient regular-level condition is $\nabla f(x)\ne 0$ at every point with $f(x)=0$. The implicit function theorem then makes the zero level a two-dimensional continuously differentiable submanifold. Its tangent space isFor a minimizing pair $(a,b)$ and any tangent vector $v\in T_a{S}_f$, take a differentiable surface curve $x(s)$ with $x(0)=a$, $x^{\prime }(0)=v$. The function $\Vert x(s)-b{\Vert }^2$ has a minimum at zero, so its derivative $2(a-b)\cdot v$ vanishes. Varying the other point similarly proves orthogonality to $T_b{S}_g$.

For the particular surfaces, minimize the distance from the origin to the second surface first. Put $u=x^2$, $v=y^2$, so $u,v\ge 0$ and $z=(u-v-6)/2$. ThenLet $s=u-v$. Since $u+v\ge |s|$, this is at least $|s|+(s-6{)}^2/4$, with equality when one of $u,v$ is zero. For $s\ge 0$, this quadratic has its minimum at $s=4$, with value 5. For $s\le 0$, it decreases up to zero and has minimum9 on that half-line. The global minimum is therefore5, attained atAll points of this surface lie outside the unit sphere. For any such $b$, the distance to the sphere is $\Vert b\Vert -1$, attained at $a=b/\Vert b\Vert$ by the triangle inequality. ConsequentlyThis is a global minimization, not merely a stationary-point calculation.