Actually, I have the problem with one part of it (condition of nonexistence of an extreme value).
Let's assume that $f(x_1, x_2, \ldots, x_n)$ is continuous and has continuous first and second derivatives in the neighbourhood of the stationary point $(x_1^0, x_2^0, \ldots, x_n^0)$.
From the Taylor theorem $$ \Delta=f(x_1, x_2, \ldots, x_n) - f(x_1^0, x_2^0, \ldots, x_n^0) = \frac{1}{2}\sum_{i,k=1}^n f_{x_ix_k}\Delta x_i \Delta x_k, $$ where $\Delta x_i = x_i - x_i^0$ and all the derivatives are taken in some point $$ (x_1^0+\theta\Delta x_1, x_2^0 + \theta \Delta x_2, \ldots, x_n^0 + \theta \Delta x_n),\quad \theta\in (0,1). $$
Let's define $$ a_{ik} = f_{x_i x_k}(x_1^0, x_2^0, \ldots, x_n^0), \quad i,k=1,2,\ldots,n. $$ From the continuity of the derivatives, we can write $$ f_{x_ix_k}(x_1^0+\theta\Delta x_1, \ldots, x_n^0 + \theta \Delta x_n) = a_{ik} + \alpha_{ik}, $$ where $\alpha_{ik}\to 0$ when $\Delta x_1 \to 0, \ldots, \Delta x_n \to 0$. Then, we can write $$ \Delta = \frac{1}{2} \left[ \sum_{i,k=1}^n a_{ik}\Delta x_i \Delta x_k + \sum_{i,k=1}^n \alpha_{ik}\Delta x_i \Delta x_k \right] $$ Now, let's assume that the first quadratic form in the [...] is indefinite. We have to show that the function $f$ doesn't have the extreme value in $(x_1^0,\ldots,x_n^0)$.
Let's assume that for some $\Delta x_i = h_i$ our form is negative $$ \sum_{i,k=1}^n a_{ik}h_i h_k <0 $$ and put $$ \Delta x_i = h_i t,\quad t\neq 0. $$ It represents moving the point $(x_1,\ldots,x_n)$ on the straight line connecting $(x_1^0, x_2^0, \ldots, x_n^0)$ and $(x_1^0 +h_1, x_2^0 +h_2, \ldots, x_n^0 + h_n)$. Then we have $$ \Delta = \frac{t^2}{2} \left[ \sum_{i,k=1}^n a_{ik}h_i h_k + \sum_{i,k=1}^n \alpha_{ik}h_i h_k \right]. $$ We have to show now that for the sufficiently small $t$, our $\Delta$ is always negative. Of course, because $t\to 0 \implies \Delta x_i \to 0 \implies \alpha_{ik}\to 0$, the second sum goes to $0$. However, for the same reason so is the first sum. How to show that the expression in the brackets [...] is always negative?
If we can show it, then it's just the matter of taking some $\Delta x_i = h_i^\ast$ for which the form is positive and the reasoning is the same. We then know that arbitrarily close to the stationary point, function $f$ takes both greater and smaller values than $f(x_{1}^0,\ldots, x_n^0)$ and thus cannot have an extreme value there.