In this situation you may immediately write the corresponding second differential of $f$:
$$d^2 f=\frac{\partial^2f}{\partial x^2}dx^2+\frac{\partial^2f}{\partial y^2}dy^2+\frac{\partial^2f}{\partial z^2}dz^2+2\left(\frac{\partial^2f}{\partial x\partial y} dx\,dy+\frac{\partial^2f}{\partial x\partial z} dx\,dz+\frac{\partial^2f}{\partial y\partial z} dx\,dy\right)$$
and check is $d^2 f$ in the given point $(x_0,\,y_0,\,z_0)$ positive definite or negative definite quadratic form.
Edited per Ryan's clarification below.
Statement 1: Yes, this is fine. If $M$ is neither positive nor negative definite, and has no zero eigenvalues, then it must have at least one positive and one negative eigenvalue. Notice that this is a sufficient but not necessary condition on $M$ being indefinite. $\left[\begin{array}{ccc}0 & 0 &0\\0 & 1 & 0\\0 & 0 & -1\end{array}\right]$ is indefinite, for instance.
Statement 2: No, this is false. Consider for instance $\left[\begin{array}{cc}1 & 0\\0 & 0\end{array}\right]$ which is positive-semidefinite.
It is impossible to characterize indefinite matrices from the leading minors alone. For example, if the first row and column of a symmetric matrix $M$ is zero, the matrix might be positive-semidefinite, negative-semidefinite, or indefinite, yet all of the leading minors will be zero.
A complete, correct statement requires looking at all principal minors, for example: a symmetric matrix $M$ is indefinite (has positive and negative eigenvalues) if and only if:
- $\Delta_k < 0$ for some even $k$; or
- $\Delta_{k_1} > 0$ and $\Delta_{k_2} < 0$ for two different odd $k_1$ and $k_2$.
Knowing that $M$ is not strictly positive- or negative-definite does not really help. You can check that if $M$ satisfies neither of these conditions, then it must satisfy one of the rows of the purple box.
EDIT: Proof of the "only if" direction. Let $M$ be indefinite. Suppose, for contradiction, that neither of the above two hold. Then either all of the odd-dimensional minors are nonnegative, or all are nonpositive.
In the former case, $M$ satisfies the third row of the purple box above, and $M$ is positive-semidefinite, a contradiction.
In the latter case, $M$ satisfies the fourth row of the purple box above, and $M$ is negative-semidefinite, a contradiction.
EDIT 3: Proof of the "if" direction. Suppose one of the even-dimensional minors is negative, and suppose, for contradiction, that $M$ is positive-semidefinite or negative-semidefinite. Then by row three or four of the purple box (as appropriate), that minor is in fact positive, a contradiction. Therefore $M$ is neither positive- nor negative-semidefinite, and so is indefinite.
Suppose instead one of the odd-dimensional minors is positive, and another is negative, and suppose $M$ is positive-semidefinite. Then both of those minors are positive, a contradiction. Now suppose $M$ is negative-semidefinite. Then both of those minors are negative, a contradiction. The only remaining possibility is that $M$ is indefinite.
Best Answer
The eigenvalues of a Hermitian matrix are real. If there is only one negative eigenvalue then the determinant, which is product of the eigenvalues would be $\leq 0$.