How does the method of Lagrange multipliers fail (in classical field theories with local constraints)?

Generically, the $m$ equations $g_i(x)=0$ define a manifold $S$ of dimension $d:=n-m$. At each point $p\in S$ the $m$ gradients $\nabla g_i(p)$ are orthogonal to the tangent space $S_p$ of $S$ at $p$. The condition rnk$(\nabla g(p))=m$ means that these $m$ gradients are linearly independent, so that they span the full orthogonal complement $S_p^\perp$ which has dimension $m=n-d$. At a conditionally stationary point $p$ of $f$ the gradient $\nabla f(p)$ is in $S_p^\perp$, and if the rank condition is fulfilled, there will be constants $\lambda_i$ such that $\nabla f(p)=\sum_{i=1}^m \lambda_i\nabla g_i(p)$. In this case the given "recipe" will find the point $p$.

Consider now the following example where the rank condition is violated: The two constraints $$g_1(x,y,z):=x^6-z=0,\qquad g_2(x,y,z):=y^3-z=0$$ define a curve $S\subset{\mathbb R}^3$ with the parametric representation $$S: \quad x\mapsto (x,x^2,x^6)\qquad (-\infty < x <\infty).$$ The function $f(x,y,z):=y$ assumes its minimum on $S$ at the origin $o$. But if we compute the gradients $$\nabla f(o)=(0,1,0), \qquad \nabla g_1(o)=\nabla g_2(o)=(0,0,-1),$$ it turns out that $\nabla f(o)$ is not a linear combination of the $\nabla g_i(o)$. As a consequence Lagrange's method will not bring this conditionally stationary point to the fore.