 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, solveThese 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.