Answer
$$\frac{{dy}}{{dx}} = \frac{{\left( {{{\tan }^{ - 1}}y + 3{x^2}} \right)\left( {1 + {y^2}} \right)}}{{{e^y}\left( {1 + {y^2}} \right) - x}}$$
Work Step by Step
$$\eqalign{
& {x^3} + x{\tan ^{ - 1}}y = {e^y} \cr
& {\text{Differentiate with respect to }}x{\text{ }} \cr
& \frac{d}{{dx}}\left[ {{x^3} + x{{\tan }^{ - 1}}y} \right] = \frac{d}{{dx}}\left[ {{e^y}} \right] \cr
& {\text{Use product rule}} \cr
& \frac{d}{{dx}}\left[ {{x^3}} \right] + x\frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}y} \right] + {\tan ^{ - 1}}y\frac{d}{{dx}}\left[ x \right] = \frac{d}{{dx}}\left[ {{e^y}} \right] \cr
& {\text{Compute the derivatives}} \cr
& 3{x^2} + x\left( {\frac{1}{{1 + {y^2}}}} \right)\frac{{dy}}{{dx}} + {\tan ^{ - 1}}y\left( 1 \right) = {e^y}\frac{{dy}}{{dx}} \cr
& {\text{Rearrange the equation}} \cr
& x\left( {\frac{1}{{1 + {y^2}}}} \right)\frac{{dy}}{{dx}} - {e^y}\frac{{dy}}{{dx}} = - {\tan ^{ - 1}}y - 3{x^2} \cr
& {\text{Solve the equation for }}\frac{{dy}}{{dx}} \cr
& \left( {\frac{x}{{1 + {y^2}}} - {e^y}} \right)\frac{{dy}}{{dx}} = - {\tan ^{ - 1}}y - 3{x^2} \cr
& \left( {\frac{{x - {e^y}\left( {1 + {y^2}} \right)}}{{1 + {y^2}}}} \right)\frac{{dy}}{{dx}} = - {\tan ^{ - 1}}y - 3{x^2} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( { - {{\tan }^{ - 1}}y - 3{x^2}} \right)\left( {1 + {y^2}} \right)}}{{x - {e^y}\left( {1 + {y^2}} \right)}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {{{\tan }^{ - 1}}y + 3{x^2}} \right)\left( {1 + {y^2}} \right)}}{{{e^y}\left( {1 + {y^2}} \right) - x}} \cr} $$
