5^(y-x)*(y+x) = 1
and
(x+y)^(x-y) = 5

assume A=log 5 to the base (x+y)

taking log to the base (x+y) eq 1 becomes

(y-x)A =-1

eq 2 becomes

(x-y)=A  substituting

(x-y)² = 1

x-y= -1 or x-y=1

substituting x-y=1 in eq 2

(2y+1) ^ 1 = 5

y=2,x=3

CHECKING :

  5^(2-3) * (3+2) = 1 in first equation

 (3+2)^(3-2)=5 hence second eqn is proved