Problem studying stationary points I have this function:$$f(x,y)=\frac{10+x^2y^2}{e^{x^2+y^2} }$$
I have found that the only stationary point is $(0,0)$ but I don't know ho to classify it as a max, min or saddle point. Can someone help me?
(I could study it using the determinant of the Hessian but I'm looking for another solution)

 A: You could taylor expand to second order $$ f(x,y) = 10e^{-(x^2+y^2)}(1+x^2y^2)\approx 10(1-(x^2+y^2))(1) =  10-10(x^2+y^2)$$
and infer that it's a local maximum (since it's an upside down paraboloid).
A: 
Method 1. One may observe that, as $(x,y) \to (0,0)$,
  $$
\begin{align}
10-f(x,y)&=10-10e^{-(x^2+y^2)}-x^2y^2e^{-(x^2+y^2)}
\\\\&=10-10e^{-(x^2+y^2)}-o(x^2+y^2)
\\\\&=10-10[1-(x^2+y^2)+o(x^2+y^2)]-o(x^2+y^2)
\\\\&=10(x^2+y^2)+o(x^2+y^2)
\end{align}
$$ which is non-negative as as $(x,y) \to (0,0)$.

Method 2. One may use the second derivative test for functions of two variables, that is evaluating
$$
D(x,y)= \frac{\partial^2 f}{\partial x^2}(x,y)\frac{\partial^2 f}{\partial y^2}(x,y) - \left(\frac{\partial^2 f}{\partial xy}(x,y) \right)^2
$$ at $(0,0)$  checking if $D(0,0)>0$ and $\frac{\partial^2 f}{\partial x^2}(0,0)<0$ to know if it is a local maximum.
A: Writing in polar form with the substitutions
$$\begin{aligned} x &= r \cos \theta & y &= r \sin \theta \end{aligned}$$
gives
$$ f = \frac{10 + r^4 (\cos \theta \sin \theta)^2}{e^{r^2}} $$
Defining $t =  (\cos \theta \sin \theta)^2$ and expanding
$$ e^{-r^2} = 1 - r^2 + \frac{1}{2}r^4 - o(r^6) $$
before multiplying gives
$$\begin{aligned}
f &= 10 - 10 r^2 + (5 + t) r^4 - o(r^6) \\
\frac{\partial}{\partial r} f &= -20 r + 4(5+t) r^3 - o(r^5) \\
\frac{\partial^2}{\partial r^2} f &= -20 + 12(5 + t)r^2 - o(r^4) \\
&< 0 &r=0, \forall \theta
\end{aligned}$$
That is, $f_\theta(r)$ has a maximum at $r = 0$ for all $\theta$.  Therefore $f(x, y)$ must have a maximum at $(x, y) = (0, 0)$.
