Critical points of $f(x,y)=x^2+xy+y^2+\frac{1}{x}+\frac{1}{y}$ I would like some help finding the critical points of $f(x,y)=x^2+xy+y^2+\frac{1}{x}+\frac{1}{y}$. I tried solving $f_x=0, f_y=0$ (where $f_x, f_y$ are the partial derivatives) but the resulting equation is very complex. The exercise has a hint: think of $f_x-f_y$ and $f_x+f_y$. However, I can't see where to use it.
Thanks!

After doing some computations I found the following (lets hope I didn't make any mistakes). You need to solve the equations
$$f_x = 2x + y - \frac{1}{x^2} = 0 \quad f_y = 2y + x -\frac{1}{y^2} = 0$$
therefore after subtracting and adding them as in the hint we get
$$\begin{align}
f_x - f_y &= x - y - \frac{1}{x^2} + \frac{1}{y^2} = 0  \\
f_x + f_y &=  3x + 3y -\frac{1}{x^2} - \frac{1}{y^2} = 0
\end{align}
$$
but you can factor them a little bit to get
$$
\begin{align}
f_x - f_y &= x - y + \frac{x^2 - y^2}{x^2 y^2} = (x - y) \left ( 1 + \frac{x+ y}{x^2 y^2}\right ) = 0\\
f_x + f_y &= 3(x + y) -\frac{x^2 + y^2}{x^2 y^2} = 0
\end{align}
$$
Now from the first equation you get two conditions, either $x = y$ or $x+y = -x^2 y^2$.
If $x = y$ you can go back to your first equation for $f_x$ and substitute to get 
$$2x + x - \frac{1}{x^2} = 0 \implies 3x = \frac{1}{x^2} \implies x = \frac{1}{\sqrt[3]{3}}$$
and then you get the critical point $\left ( \dfrac{1}{\sqrt[3]{3}}, \dfrac{1}{\sqrt[3]{3}} \right )$
Now if instead $x + y = -x^2 y^2$ then if you substitute into the equation $f_x + f_y = 0$ we get the following
$$
3(-x^2 y^2) - \frac{x^2 + y^2}{x^2 y^2} = 0 \implies 3x^4 y^4 + x^2 + y^2 = 0 \implies x = y = 0 
$$
But this is actually one of the points where the partial derivatives or even your original function are not defined.
