Answer
$$\frac{{dy}}{{dx}} = - \frac{{y + y\sec y}}{{3x + 4y{{\left( {1 + \sec y} \right)}^2} + 3x\sec y - xy\sec y\tan y}}$$
Work Step by Step
$$\eqalign{
& \frac{{x{y^3}}}{{1 + \sec y}} = 1 + {y^4} \cr
& {\text{Differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left[ {\frac{{x{y^3}}}{{1 + \sec y}}} \right] = \frac{d}{{dx}}\left[ {1 + {y^4}} \right] \cr
& \frac{{\left( {1 + \sec y} \right)\left( {3x{y^2}y' + {y^3}} \right) - x{y^3}\left( {y'\sec y\tan y} \right)}}{{{{\left( {1 + \sec y} \right)}^2}}} = 4{y^3}y' \cr
& \left( {1 + \sec y} \right)\left( {3x{y^2}y' + {y^3}} \right) - x{y^3}\left( {y'\sec y\tan y} \right) = 4{y^3}y'{\left( {1 + \sec y} \right)^2} \cr
& {\text{Simplifying}} \cr
& 3x{y^2}y' + {y^3} + 3x{y^2}\sec yy' + {y^3}\sec y - x{y^3}\sec y\tan yy' = 4{y^3}y'{\left( {1 + \sec y} \right)^2} \cr
& 3x{y^2}y' + 4{y^3}y'{\left( {1 + \sec y} \right)^2} + 3x{y^2}\sec yy' - x{y^3}\sec y\tan yy' = - {y^3} - {y^3}\sec y \cr
& {\text{Factoring}} \cr
& \left[ {3x{y^2} + 4{y^3}{{\left( {1 + \sec y} \right)}^2} + 3x{y^2}\sec y - x{y^3}\sec y\tan y} \right]y' = - {y^3} - {y^3}\sec y \cr
& {\text{Solving for }}y' \cr
& y' = \frac{{ - {y^3} - {y^3}\sec y}}{{3x{y^2} + 4{y^3}{{\left( {1 + \sec y} \right)}^2} + 3x{y^2}\sec y - x{y^3}\sec y\tan y}} \cr
& \frac{{dy}}{{dx}} = - \frac{{y + y\sec y}}{{3x + 4y{{\left( {1 + \sec y} \right)}^2} + 3x\sec y - xy\sec y\tan y}} \cr} $$
