Find $\frac {dy}{dx}$ in the following: $ax +by^2 = cos \: y$

Question

$\begin{array}{1 1} \large \frac{-a}{siny+2yb} \\ \large \frac{a}{siny+2yb} \\\large \frac{-a}{-siny+2yb} \\\large \frac{-a}{-siny+2yb} \end{array} $