Answer
$$\frac{{dy}}{{dx}} = - \frac{{2xy}}{{2{y^2} + \left( {{x^2} + {y^2} + 4} \right)\ln \left( {{x^2} + {y^2} + 4} \right)}}$$
Work Step by Step
$$\eqalign{
& y\ln \left( {{x^2} + {y^2} + 4} \right) = 3 \cr
& {\text{Let }}F\left( {x,y} \right) = y\ln \left( {{x^2} + {y^2} + 4} \right) - 3 = 0 \cr
& {\text{Calculate the partial derivatives }}{F_x}\left( {x,y} \right){\text{ and }}{F_y}\left( {x,y} \right) \cr
& {F_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {y\ln \left( {{x^2} + {y^2} + 4} \right)} \right] \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = y\left( {\frac{{2x}}{{{x^2} + {y^2} + 4}}} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2xy}}{{{x^2} + {y^2} + 4}} \cr
& {F_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {y\ln \left( {{x^2} + {y^2} + 4} \right)} \right] \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = y\left( {\frac{{2y}}{{{x^2} + {y^2} + 4}}} \right) + \ln \left( {{x^2} + {y^2} + 4} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2{y^2}}}{{{x^2} + {y^2} + 4}} + \ln \left( {{x^2} + {y^2} + 4} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2{y^2} + \left( {{x^2} + {y^2} + 4} \right)\ln \left( {{x^2} + {y^2} + 4} \right)}}{{{x^2} + {y^2} + 4}} \cr
& {\text{Therefore,}} \cr
& \frac{{dy}}{{dx}} = - \frac{{{F_x}}}{{{F_y}}} = - \left( {\frac{{2xy}}{{{x^2} + {y^2} + 4}}} \right)\left( {\frac{{{x^2} + {y^2} + 4}}{{2{y^2} + \left( {{x^2} + {y^2} + 4} \right)\ln \left( {{x^2} + {y^2} + 4} \right)}}} \right) \cr
& {\text{Simplifying}} \cr
& \frac{{dy}}{{dx}} = - \frac{{2xy}}{{2{y^2} + \left( {{x^2} + {y^2} + 4} \right)\ln \left( {{x^2} + {y^2} + 4} \right)}} \cr} $$
