problem: $\displaystyle 1+x = sin(xy^2) $

solution attempt: differentiating both sides, I have

$\displaystyle 1 = -cosx \cdot\frac{d}{dx}[xy^2] $

$\displaystyle 1 = -cosx(y^2+2xyy')$

dividing by $\displaystyle -cosx$ on both sides, I have

$\displaystyle -\frac{1}{cosx} =y^2+2xyy'$

dividing by $\displaystyle 2xy$ on both sides, I have

$\displaystyle -\frac{2xy}{cosx} = \frac{y^2}{2xy} + y'$

subtracting $\displaystyle \frac{y^2}{2xy}$ from both sides, I have

$\displaystyle y' = -\frac{2xy}{cosx}-\frac{y^2}{2xy}$

then,

$\displaystyle y' = -\frac{y^2(4x^2 - cosx)}{2xycosx}$