$ax + b{y^2} = \cos y$

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

$a + b\left[ {2y\cfrac{{dy}}{{dx}}} \right] = - \sin y\cfrac{{dy}}{{dx}} \Rightarrow a + 2by\cfrac{{dy}}{{dx}} + \sin y\cfrac{{dy}}{{dx}} = 0$

$\Rightarrow$ $\cfrac{{dy}}{{dx}}[2by + \sin y] = - a \Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{{ - a}}{{2by + \sin y}}$