If $\tan y+y=secx$ then find $\frac{dy}{dx}$.

$\tan y+y=\sec x$

$\Rightarrow {\sec }^{2}y\frac{dy}{dx}+\frac{dy}{dx}=\sec x\tan x$

$\Rightarrow ({\sec }^{2}y+1)\frac{dy}{dx}=\sec x\tan x$

$\therefore \frac{dy}{dx}=\frac{\sec x\tan x}{({\sec }^{2}y+1)}$