Why does the second derivative test in multivariable functions work? I haven't seen this asked much, if at all, and a lot of college books in my knowledge never provide the reasoning behind why the second derivative test even works?
$$\frac{\partial^2 z}{\partial x^2}\cdot\frac{\partial^2 z}{\partial y^2} - \left(\frac{\partial^2 z}{\partial x \partial y}\right)^2$$
If the value of that is less than $0$, then the critical point is a saddle point, if the value is more than $0$ then the critical point is either a maximum or a minimum. Otherwise the test is inconclusive.
I just wanted to ask why is that? Can anyone give me good intuition on why that is the case?
 A: The reason why regular college textbooks don't justify why this works, is because the proof is too advanced. You cannot get 20 year old college kids, who are taking their 3rd course in Calculus, to truly understand it. You need to dig deeper and check out Real Analysis textbooks. Find a Real Analysis textbook, go to the multivariate part, and you'll find a proof.
A hint on maybe finding some intuitive answer about why this works: on YouTube there's a playlist called Khan Academy multivariate calculus which is taught by Grant Sanderson, the man behind the 3Blue1Brown YouTube channel. Search for the video about the second derivative test, and you will probably have your answer. Grant likes to give intuitive explanations about why things are the way they are instead of just hammering the proof on your soul.
A: Assume that $f:\>{\mathbb R}^2\to{\mathbb R}$ is twice differentiable. When $\nabla f(x_0,y_0)=(0,0)$ then $(x_0,y_0)$ is a critical point of $f$. Taylor's theorem then says that when $(X,Y)\to(0,0)$ we have
$$f(x_0+X,y_0+Y)=f(x_0,y_0)+aX^2+2bXY+cY^2+o(X^2+Y^2)\ ,\tag{1}$$
whereby
$$a={1\over2}f_{xx}(x_0,y_0),\quad b={1\over2}f_{xy}(x_0,y_0),\quad c={1\over2}f_{yy}(x_0,y_0)\ .$$
The behavior of $f$ in the neighborhood of $(x_0,y_0)$ is completely determined by the three values $a$, $b$, $c$, resp., by the quadratic form
$$q(X,Y):=aX^2+2bXY+cY^2\ .$$
In linear algebra it is shown tht such a quadratic form can be positive definite, negative definite, indefinite, or degenerate. In the first three cases the "second derivative test" gives a factual result. The case we have before us is determined by the signs of $a$ and $ac-b^2$.
As an example we consider the case $a>0$, $ac-b^2>0$. In this case the form $q$ is positive definite, i.e., we have $q(X,Y)>0$ for all $(X,Y)\ne(0,0)$. One then can show that there is a $\lambda>0$ with
$$q(X,Y)\geq\lambda(X^2+Y^2)\ .$$
From $(1)$ we then obtain
$$f(x_0+X,y_0+Y)-f(x_0,y_0)\geq {\lambda\over2}(X^2+Y^2)>0\qquad\bigl(0<\sqrt{X^2+Y^2}<\delta\bigr)\ ,\tag{2}$$
whereby $\delta>0$ has been chosen such that $|o(X^2+Y^2)|<{\lambda\over2}(X^2+Y^2)$ when $\sqrt{X^2+Y^2}<\delta$.
The equation $(2)$ shows that $f$ has a "strong" local minimum at $(x_0,y_0)$.
