If a function of two variables has a unique critical point, which is a local maximum, is it a global maximum? 
$f(x,y)$ has partial derivatives in all $\mathbb R^2$ and a unique critical point at $(x_0,y_0)$ (local maximum). Is it a global maximum?

I know that in compact sets, it isn't enough to say that if a point is the only maximum inside the set, then it's a global maximum, because in the frontier of the set it could happen that the function has a maximum greater than the inside one. But for the entire $\mathbb R^2$ there is no frontier, therefore can I admit that this unique point is a point of global maximum?
 A: Loose description of the geometry: imagine a flat plane and then you put a lone hill with a peak on it. Now tilt the plane a little. Now you still have one peak, but hopefully you can also see that you have introduced a saddle point. Imagine sliding the saddle point location off to infinity to make it effectively no longer there.
Basic description of the construction I saw somewhere once: you can have something whose cross sections in one direction are cubic, but whose cross-sections in the other direction is a sum of exponentials in just the right way for one of the cubic cross-sectional peaks to meet the peak of one of the exponential sum cross-sections. This can happen with no other critical points, and of course the cubic component alone is unbounded in either direction (so there is no global max or min). The direction that the exponentials taper off in aligns with the idea of the saddle point being slid away to infinity.
Consider $f(x,y)=x^3+axe^{my}+be^{ny}$ for some constants $a,b,m,n$. We have:
$$\begin{align}
\frac{\partial}{\partial x}f(x,y)&=3x^2+ae^{my}&\frac{\partial}{\partial y}f(x,y)&=amxe^{my}+bne^{ny}
\end{align}$$
To find a critical point, the $\frac{\partial}{\partial x}$ shows us $a$ must be negative. We are just looking for one example, so we pick some numbers for the constants that make the algebra that follows simpler. With $a=-3$, $b=1$, $m=1$, $n=3$:
$$\begin{align}
\frac{\partial}{\partial x}f(x,y)&=3x^2-3e^{y}&\frac{\partial}{\partial y}f(x,y)&=-3xe^{y}+3e^{3y}\\
&=3(x^2-e^y)&&=3e^y(e^{2y}-x)
\end{align}$$
So at a critical point, $x=e^{2y}$ according to the latter relation.
Then using the $\frac{\partial}{\partial x}$ relation:
$$e^{4y}-e^y=0\implies e^{3y}=1\implies y=0$$
which in turn says that $x=1$. So the only critical point is at $(1,0)$. Is it a local min, max, or neither?
$$\begin{align}
f_{xx}(1,0)&=6&f_{xy}(1,0)&=-3&f_{yy}=6
\end{align}$$
So the Hessian determinant is $6^2-(-3)^2=27>0$, so this critical point is either a local max or a local min. OK, it turns out to be a local min, since for example $f_{xx}(1,0)$ is positive. But $f$ has no global min, since you could fix $y$ and let $x\to-\infty$ and get unbounded large negative output.
Take this $f$ and negate it to get the direct counterexample. Here is an image of $-f$ from GeoGebra. Note how there would be a saddle point, except it has been slid off the map (to infinity) to the left.

