Determine all local minimums, maximums and saddle points for $f(x,y)= x^2y +4/x +4/y$ $$f_x=2xy-\frac{4}{x^2}$$ $$f_y=x^2-\frac{4}{y^2}$$
And now when i'm trying to solve that i get somethink like this
$$ x^2-x^6=0$$ $\to x=0, x=1, x=-1$ and when i'm putting this into this equation
$$ y=\frac{2}{x^3}$$
i get $y=2$ or $y=-2$. But what's  next? I need to shuffle it one by one getting 6 points $(1,2), (1,-2), (0,2), (0,-2), (-1,2), (-1,-2)$? Or i just need to get that points that match. I mean $(1,2), (-1,-2)$.
I know that can be really stupid question but after my summer pause my brain is in $90%$ dead. The next steps are trivial but i don't know which points i need to get :D

$$2 x y-\frac{4}{x^2}=0,x^2-\frac{4}{y^2}=0$$
has $(-1,-2)$ and $(1,2)$ as solutions. ($x=0$ cannot be a solution since $0$ is not in the domain of the function)
Calculate the second derivatives we get the Hessian
$$H(x,y)=\det\left(
\begin{array}{ll}
 2 y+\dfrac{8}{x^3} & 2 x \\
 2 x & \dfrac{8}{y^3} \\
\end{array}
\right)=\frac{64}{x^3 y^3}-4 x^2+\frac{16}{y^2}
$$
$H(-1,-2)=8;\dfrac{\partial^2 f}{\partial x^2}\,(-1,-2)=-12$
So $(-1;\;-2)$ is a maximum, while
$H(1,2)=8;\dfrac{\partial^2 f}{\partial x^2}\,(1,2)=12$
So $(1;\;2)$ is a minimum.
For further details look here
