Find $ \frac {dy}{dx} $ in the following: $ xy +y^2 = tan \: x + y $

Question

$\begin{array}{1 1} \frac{sec^2x-y}{x+2y-1} \\ \large \frac{sec^2x+y}{x+2y-1} \\ \frac{cosec^2x-y}{x+2y-1} \\ \frac{cosec^2x+y}{x+2y-1} \end{array} $