Calculate the partial derivative, local minima/maxima, and saddle points. 
I am having trouble finding the partial derivative. And clues or hints regarding said problem and how to find saddle points/local maxima is appreciated.
 A: Partial derivatives are calculated by regarding the function as a function in only one argument and considering the other variables as constants.
So you obtain $f_x(x,y)=\frac{d}{dx}(y x -2 x^{-1}-2/y)=y+2x^{-2}$.
Saddle points and local maxima/minima are always at places where both derivatives vanish simultaneously. To decide which is which, you have to look at the second derivatives:


*

*$f_x(x_0,y_0)>0,f_y(x_0,y_0)>0$: $(x_0,y_0)$ is local minimum

*$f_x(x_0,y_0)<0,f_y(x_0,y_0)<0$: $(x_0,y_0)$ is local maximum

*$f_x(x_0,y_0)>0,f_y(x_0,y_0)<0$ or$f_x(x_0,y_0)<0,f_y(x_0,y_0>0$: $(x_0,y_0)$ is a saddle point.

A: I often find that the best way to go about optimization problems is to go about mapping the derivatives out, and then plugging in points:


*

*fx=y+2/(x^2)

*fxx=-4/(x^3)

*fy=x+2/(y^2)

*fyy=-4/(y^3)

*fxy=0



Now you plugin the desired points into your new equations and evaluate. Values wherein fx and fy are equal to 0 are said to be critical points. These are the extrema you are looking for and can be classified using the following equation:


*

*D=[(fxx)(fyy)]-[(fxy)^2]

If D is positive and both fxx and fyy are negative at the chosen values, then the critical point is a maximum. If D  is positive and both fxx and fyy are positive at the chosen values, then the critical point is a minimum. If D is negative at the chosen values, then the critical point is a saddle. If D is zero then the nature of the critical point is indeterminable, it could be any kind of point, geometric analysis would be necessary.
