simplest way here is to convert this to a problem in **single variable calculus**

#f(x,y)=(x-y)/x^2# and #xy = 4#

#implies f(x)=(x-4/x)/x^2 = 1/x - 4/x^3#

#f'(x) = - 1/x^2 + 12/x^4#

#f'(x) = 0 implies 12/x^4 - 1/x^2 = 0#

# 12 - x^2 = 0, x = pm 2 sqrt 3#

and #y = 4/x = pm 4/( 2 sqrt 3) = pm 2/sqrt3#

so the fixed points are

#(2 sqrt 3, 2/sqrt3) # and #(-2 sqrt 3, -2/sqrt3) #

the advantage of single var over Lagrange Multipe is that the nature of the fixed points are found mechanically via the second derivative

#f''(x) = 2/x^3 - 48/x^5#

#= 2/x^3 ( 1 - 24/x^2) #

#x = + 2 sqrt 3, f'' = 2/(2 sqrt 3)^3 (1- 2)# ie negative, so this is a local max

#x = - 2 sqrt 3, f'' = -2/(2 sqrt 3)^3 (1- 2)# ie positive, so this is a local min

this is consistent with evaluation of the function at these points

#f_max = f( 2 sqrt 3) = 1/(3 sqrt 3)#

#f_min = f( -2 sqrt 3) = -1/(3 sqrt 3) #

repeating this to check using the **Lagrange Multiplier** to optimise function #f(x,y)=(x-y)/x^2 = 1/x - y / x^2# subject to constraint #g(x,y) = xy - 4 = 0#

#nabla f = lambda nabla g#

#implies ((- 1/x^2+ (2y)/x^3),(- 1/x^2)) = lambda ((y),(x))#

#implies \lambda = (- 1/x^2+ (2y)/x^3)/(y) = (- 1/x^2)/(x)#

#implies - 1/x+ (2y)/x^2 = - y/x^2#

#implies (3y)/x^2 = 1/x, 3y = x# telling us that the critical points lie on the intersection of #xy = 4# and #y = 1/3 x#

so if we plug this into the constraint #xy = 4# then

#3y*y = 4, y = pm 2/(sqrt 3)# same as above

further work on the Hessian matrix might reveal the nature of the fixed points but that is already apparent from the single variable approach