Answer:
Explanation:
Given is a function f in two variables as

To find critical points of the given function and also to find max, min or saddle point.
Find the partial derivatives



Equate first derivatives to 0
Adding fx and fy we get

x=±y
Only real roots are (x,y) =(3,3) and (-3,-3)

(3,3) is a saddle point

(-3,-3) is also a saddle point.