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

Hint
See that $(0,0)$ is not in the domain. You then got, $x\ne0$ and $y\ne0$:
$$xy=\frac{2}{x^2}\to (xy)^2=\frac{4}{x^4}\quad (1)$$
and
$$x^2=\frac{4}{y^2}\to (xy)^2=4\quad (2)$$
and then
$$4=\frac{4}{x^4}\to x=\pm1$$
and backing to $(1)$ we get $y=\pm2$ what give us the pairs $(1,2)$ and $(-1,-2)$ as candidates.
Can you finish?
