Prove that an isolated point of $C: (f=0)\subset \mathbb{R}^2$ must be a max or min of $f: \mathbb{R}^2 \rightarrow \mathbb{R}$. Let $f\in \mathbb{R}[x,y]$ and let $C: (f=0)\subset \mathbb{R}^2$; we say that $P\in C$ is isolated if there is an $\epsilon >0$ such that $C\cap B(P,\epsilon)=P$. Prove that if $P\in C$ is an isolated point then $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ must have a max or min at $P$, and deduce that $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ vanish at $P$. 
I can imagine how it is. Since $f$ is a polynomial, it is a smooth surface in $\mathbb{R}^3$. If its intersection with the plane $z=0$ is an isolated point, it can only touch $z=0$ at that point without passing through. 
So there should be a neighborhood around $P$ such that $f(x,y)>0$ or $f(x,y)<0$ in that neighborhood. The second part is then trivial.  
But I have no idea how to rigorously prove that it is a max or min. Thank you for your help! 
 A: Restrict your function to $B(P,\epsilon)$, such that $f^{-1}(0)=\{P\}$. Now if the partial derivatives don't vanish at $P$, you conclude that $0$ is a regular value of $f$ and hence $f^{-1}(0)$ is a smooth 1-manifold which contradicts the fact that it equals $\{P\}$. Now you know that $P$ is a critical point. Use the intermediate value theorem along paths to show that $f$ does not change sign on $B(P,\epsilon)$ and the statement follows.
EDIT: I think this statement can be found in differential geometry or differential topology (search for the theory of regular values and you should find references). What I mean by the intermediate value theorem along paths is the following: Suppose there exist points $p,q\in B(P,\epsilon)$ with $f(p)>0>f(q)$. Connect $p$ and $q$ with a continuous path $\gamma:[0,1]\to B(P,\epsilon)\setminus\{P\}$ such that $\gamma(0)=p$ and $\gamma(1)=q$. Then the map $F=f\circ \gamma :[0,1]\to \mathbb R$ is continuous and $F(0)>0>F(1)$. By the intermediate value theorem there exists $t_0\in(0,1)$ such that $f(\gamma(t_0))=0$. This means that $\gamma(t_0)\ne P$ is another point in $f^{-1}(0)$ which contradicts the fact that $f^{-1}(0)=\{P\}$. Hence $f$ does not change sign on $B(P,\epsilon)$. I hope this is clearer now, otherwise let me know.
