Answer
$$\,\,\,\,\,\,\,\,\,\,\,\,\,y = - \frac{1}{6}x + \frac{3}{2}$$
Work Step by Step
$$\eqalign{
& y = \frac{{4x}}{{{x^2} + 3}},{\text{ point }}\left( {3,1} \right) \cr
& {\text{differentiate }}y \cr
& y'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{4x}}{{{x^2} + 3}}} \right] \cr
& {\text{using the quotient rule}} \cr
& y'\left( x \right) = \frac{{\left( {{x^2} + 3} \right)\left( {4x} \right)' - 4x\left( {{x^2} + 3} \right)'}}{{{{\left( {{x^2} + 3} \right)}^2}}} \cr
& y'\left( x \right) = \frac{{\left( {{x^2} + 3} \right)\left( 4 \right) - 4x\left( {2x} \right)}}{{{{\left( {{x^2} + 3} \right)}^2}}} \cr
& {\text{simplify}} \cr
& y'\left( x \right) = \frac{{4{x^2} + 12 - 8{x^2}}}{{{{\left( {{x^2} + 3} \right)}^2}}} \cr
& y'\left( x \right) = \frac{{12 - 4{x^2}}}{{{{\left( {{x^2} + 3} \right)}^2}}} \cr
& {\text{find the slope at the point }}\left( {3,1} \right) \cr
& \,\,\,m = y'\left( 3 \right) \cr
& \,\,\,m = \frac{{12 - 4{{\left( 3 \right)}^2}}}{{{{\left( {{{\left( 3 \right)}^2} + 3} \right)}^2}}} \cr
& \,\,\,m = \frac{{ - 24}}{{144}} \cr
& \,\,\,m = - \frac{1}{6} \cr
& {\text{find the equation of the tangent line at the point }}\left( {3,1} \right) \cr
& \,\,\,\,\,y - {y_1} = m\left( {x - {x_1}} \right) \cr
& \,\,\,\,\,y - 1 = - \frac{1}{6}\left( {x - 3} \right) \cr
& {\text{simplify}} \cr
& \,\,\,\,\,y - 1 = - \frac{1}{6}x + \frac{1}{2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,y = - \frac{1}{6}x + \frac{3}{2} \cr} $$
