Determine all local minimums, maximums and saddle points for $f(x,y)= x^2y +4/x +4/y$

Question

$$f_x=2xy-\frac{4}{x^2}$$ $$f_y=x^2-\frac{4}{y^2}$$

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

Answer 1

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?

Answer 2

$$2 x y-\frac{4}{x^2}=0,x^2-\frac{4}{y^2}=0$$ 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