Let $f$ and $g$ be $\mathbb{R}$-valued $C^{\infty}$ functions on $\mathbb{R}^2$ and let $S=\{(x,y)\in\mathbb{R}^2|f(x,y)=0\}$.
Suppose that at some point $p=(a,b)\in S$ we have
$\frac{\partial f}{\partial x}(p)=-1$, $\frac{\partial f}{\partial y}(p)=2$, $\frac{\partial g}{\partial x}(p)=3$, $\frac{\partial g}{\partial y}(p)=-6$,
$\begin{bmatrix} \frac{\partial^2 f}{\partial x^2}(p) & \frac{\partial^2 f}{\partial x\partial y}(p) \\ \frac{\partial^2 f}{\partial y\partial x}(p) & \frac{\partial^2 f}{\partial y^2}(p) \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 3 & 0 \end{bmatrix}$, and $\begin{bmatrix} \frac{\partial^2 g}{\partial x^2}(p) & \frac{\partial^2 g}{\partial x\partial y}(p) \\ \frac{\partial^2 g}{\partial y\partial x}(p) & \frac{\partial^2 g}{\partial y^2}(p) \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ -1 & 2 \end{bmatrix}$.
Show that there exists $R>0$ such that $g(p)<g(q)$ for $q\in S\cap\{(x,y)\in\mathbb{R}^2|(x-a)^2+(y-b)^2<R^2\}$.
I am very very sorry that I keep haven't figured out it is because what concepts/topics/applications/skills that I am unfamiliar with so that I fail to have an idea how to solve this problem from entrance exams into math Master's program in Taiwan.
I really need someone can guess what I lack of and point out the direction for me to solve this problem.
Thanks eveyone in advance for every possible help!!!
P.S. This problem already had appeared in the past exams in year 2017. It is already a past/old problem now. An past/old exam file can be downloaded from: http://www.math.ntu.edu.tw/sites/default/files/imce/documents/exams/M_aca_107.pdf
Naively this seems to test you the Hessian test for local minimum applied to the function $3f+g$. But the test indicates that the Hessian determinant $D(a,b)<0$ (using notation in the link), i.e. $(a,b)$ corresponds to a saddle point. So one needs to modify this approach. The following is a brief sketch: Without loss of generality, assume $p=(a,b)=(0,0)$ and $g(p)=0$. Then by the given assumptions, the tangent line at $p$ of the curve $f(x,y)=0$ has slope $$\frac{dy}{dx}(0,0)=-\frac{f_x(0,0)}{f_y(0,0)}=\frac 12$$ and $$3f+g=3x^2+8xy+y^2+{\rm HOT},$$ where HOT means higher order terms.
Now when $(x,y)$ approaches $(0,0)$ along $S$ in a neighborhood of $(0,0)$, one has $$y=\frac 12x+o(|x|)$$ and therefore $$g=3f+g=\frac{29}4x^2+o(x^2)$$ is strictly positive in a certain punctured disk at $(0,0)$, i.e. $g(q)>0=g(p)$ in that disk.