${(\cos x)^y} = {(\cos y)^x}$

We have, ${(\cos x)^y} = {(\cos y)^x}$
By taking log on both sides , we get

y log (cos x) $=$ x log (cos y) …(i)
Differentiating (i) on both sides w.r.t. x, we get
$\Rightarrow$ $y\cfrac{d}{{dx}}\log (\cos x) + \log (\cos x)\cfrac{{dy}}{{dx}}$

$= x\cfrac{d}{{dx}}(\log (\cos y)) + \log (\cos y)$

$\Rightarrow$ $y\cfrac{1}{{\cos x}}( - \sin x) + \log (\cos x)\cfrac{{dy}}{{dx}}$

$= x \cdot \cfrac{1}{{\cos y}}( - \sin xy)\cfrac{{dy}}{{dx}} + \log (\cos y)$
$\Rightarrow$ $\cfrac{{dy}}{{dx}}[\log (\cos x) + x\tan y] = \log (\cos y) + y\tan x$

$\Rightarrow$ $\cfrac{{dy}}{{dx}} = \cfrac{{\log (\cos y) + y\tan x}}{{\log (\cos x) + x\tan y}}$