Let $F \colon \mathbb R^2 \to \mathbb R$ be the function $ F(x,y):=xye^x + ye^y - e^x+1 $ and denote with $C$ the set of zeroes of $F$, i.e. $C:=\{(x,y) \in \mathbb R^2 : F(x,y)=0\}$. Let also $f \colon \mathbb R^2 \to \mathbb R$ be a function which is $C^2$ in a neighbourhood of $(0,0)$ and such that $ \nabla f(0,0) = (-2,2), \qquad \qquad H_f(0,0) = \begin{pmatrix} 2 & 1 \\ 1 & 0 \end{pmatrix}. $ Question: is $(0,0)$ a minimum for $f$ on $C$?
Well, I show you what I have done so far. First of all, some routine calculations yield $ \nabla F(0,0) = (-1,1) \qquad \qquad H_F(0,0) = \begin{pmatrix} -1 & 1 \\ 1 & 2 \end{pmatrix} $
Indeed, the fact that $ \nabla f(0,0) = 2\nabla F(0,0) $ does agree with the theory of Lagrange multipliers: the gradients are parellel, so I think that $(0,0)$ is an extremum for $f$ on $C$. The problem is how to classify it without any information on $f$: we have only its hessian matrix, which is - I suppose - the key to solve this. Both $H_f(0,0)$ and $H_F(0,0)$ are indefinite.
How can we establish the nature of the critical point $(0,0)$? Thanks in advance for your help.