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