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
