Given, (cos x)^y = (sin y)^x

Take log on both side, we get

y * log cos x = x * log sin y

Now, differentiate both side w.r.t. x, we get

     (dy/dx) * log cos x + y * (-sin x/cos x)=  log sin y + (cos y/sin y) * (dy/dx)

=> (dy/dx) * [log cos x - cot y] = log sin y + y * tan x

=> dy/dx =  (log sin y + y * tan x)/(log cos x - cot y)