Write , , and let , and be evaluated at the critical point. Assuming is twice continuously differentiable nearby, the Taylor expansion is
The linear terms vanish because the gradient is zero. For , completing the square gives
Consequently the requested constant is the Hessian determinant
The displayed identity in the question is an identity for the quadratic part; for a general , its right-hand side needs the remainder .
If , the quadratic form is positive for every nonzero . On the unit circle it has a positive minimum, so it dominates the remainder sufficiently near the critical point. Hence the point is a strict local minimum.
If , along the quadratic term is , whereas along it is . Along each line the remainder is of smaller order. Thus values both above and below occur arbitrarily nearby, giving a saddle point.

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