We know that Lagrange Multiplier gives necessary conditions for an extremum.It locates all possible condidates.But not all such points need be extrma.
I want to find an example of the point is neither local maxima nor local minima,but It is well satisfied the Lagrange Multiplier Condition.
In 1D calculus, an example of a function $f : \mathbb R \to \mathbb R$ where $f'(x) = 0$ does not give a local max or min is $f(x) = x^3$. Inspired by that, try this example:
Let $g(x,y) = x^2y$ with the constraint $h(x,y) = x - y = 0$. Then $\nabla g - \lambda \nabla h = 0$ iff
$$2xy - \lambda = 0 \ \text{ and } x^2 + \lambda = 0$$
I.e., $2xy = -x^2$ or $x(2y + x) = 0$. Which with the constraint necessarily implies $x = y = 0$. However $g(0,0) = 0$ is neither a local minimum or maximum.
For an example in higher dimensions, using the same inspiration:
Let $f(x_1, x_2, \cdots, x_n) = x_n$ with constraint $g(x_1, x_2, \cdots, x_n) = x_1^3 - x_n = 0$.
$\nabla f - \lambda \nabla g = 0$ iff $0 = -2\lambda x_1^2 = 1 + \lambda$ iff $\lambda = -1, x_1 = 0$. Thus $x_n = 0$, $f(x_1, ..., x_n = 0) = 0$. Which is again neither a local min or max.