$xy + {y^2} = \tan x + y$

We are given that, $xy + {y^2} = \tan x + y$ ...(i)
Differentiating (i) on both sides w.r.t. x, we get

$x\cfrac{{dy}}{{dx}} + y + 2y\cfrac{{dy}}{{dx}} = {\sec ^2}x + \cfrac{{dy}}{{dx}}$

$\Rightarrow$ $x\cfrac{{dy}}{{dx}} + 2y\cfrac{{dy}}{{dx}} - \cfrac{{dy}}{{dx}} = {\sec ^2}x - y$

$\Rightarrow$ $\cfrac{{dy}}{{dx}}[x + 2y - 1] = {\sec ^2}x - y \Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{{{{\sec }^2}x - y}}{{x + 2y - 1}}$