= Solution
Introduce the second constraint $q(x,y,z)=y-h(x)=0$ and a second <Lagrange multiplier> $\mu$. At points where the two constraint <gradients> are <linearly independent>, solve
$$
\boxed{(y,x+z,y)^T
=\lambda(2x,2y,2z)^T+\mu(-h'(x),1,0)^T},
$$
$$
\boxed{x^2+y^2+z^2=1,\qquad y=h(x)}.
$$
These are five scalar equations for $x,y,z,\lambda,\mu$. The two constraint <normal vectors> span the normal <plane> to the feasible curve. Points where those <gradients> become dependent must be examined separately; the regular two-multiplier condition need not cover them.
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