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)
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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.
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 $(x, y) = (0, 0)$.