how to Differentiate x^y+y^x=1 ?

Implicitly:

d/dx (x^y) + d/dx (y^x) = d/dx (1) = 0

d/dx (y ln x) * x^y + dy/dx * x y^(x-1) = 0
(dy/dx ln x + y/x) x^y + dy/dx * x y^(x-1) = 0

Now combine terms to separate out dy/dx and solve:
(dy/dx ln x + y/x) x^y + dy/dx * x y^(x-1) = 0
dy/dx (ln x) (x^y) + y x^(y-1) + x y^(x-1) dy/dx = 0
dy/dx (x y^(x-1) + x^y ln x) = -y x^(y-1)

dy/dx = -y x^(y-1) / (x y^(x-1) + x^y ln x)

Or in another form:
dy/dx = -y / (x^(2-y) y^(x-1) + x ln x)

Let u = x^y.
lnu = ylnx
dlnu/dx = dylnx/dx
1/u * du/dx = lnx * dy/dx + y * 1/x
du/dx = ulnx * dy/dx + uy/x

Let v = y^x.
lnv = xlny
dlnv/dx = dxlny/dx
1/v * dv/dx = x * 1/y * dy/dx + lny
dv/dx = vx/y * dy/dx + vlny

x^y + y^x = 1
ulnx * dy/dx + uy/x + vx/y * dy/dx + vlny = 0
dy/dx(ulnx + vx/y) = -uy/x - vlny
dy/dx = (-uy/x - vlny)/(ulnx + vx/y)
dy/dx = (-x^y * y/x - y^x * lny)/(x^y * lnx + y^x * x/y)