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

Let us substitute x-y = z, so that: x = y+z. Then, we must have:

(2y + z)*5^(-z) = 1 ......(i)
(2y+z)^z = 5 .....(ii)

From (ii), we have: 2y+z = 5^(1/z), and substituting this in (i), we obtain:

5^(1/z - z) = 1, so that:
1/z = z, giving: z^2 = 1, so that:
z = +/- 1

If z = 1, then (i) gives: 2y + 1 = 5, so that: y= 2, giving: x = 2 + 1 = 3

If z = -1, then (i) gives: 2y - 1 = 1/5, so that: y = 3/5, giving: x = -2/5

Consequently, the required solution is given by (x, y) = (3, 2), or (-2/5, 3/5)