$\text{(cos x)}^{\text{y}}=\text{(cos y)}^{\text{x}}\Rightarrow$ y log cos x = x log cos y.$\therefore\text{y}.\frac{\text{(-sin x)}}{\text{cos x}}+\text{log cos x.}\frac{\text{dy}}{\text{dx}}=\text{x}.\frac{\text{(-sin y)}}{\text{cos y}}\frac{\text{dy}}{\text{dx}}+\text{log cos y.}$
(log cos x + x tan y) $\frac{\text{dy}}{\text{dx}}$ = log cos y + y tan x
$\therefore\frac{\text{dy}}{\text{dx}}= \frac{\text{log cos y + y tan x}}{\text{log cos x + x tan y}}$.
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