Answer
$y = \frac{3}{4}x - \frac{1}{4}$
Work Step by Step
$$\eqalign{
& y = \frac{{{x^2}}}{{1 + x}}{\text{ at the point }}\left( {1,\frac{1}{2}} \right) \cr
& {\text{Differentiate }}y{\text{ to calculate the slope at the point }}\left( {1,\frac{1}{2}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{{{x^2}}}{{1 + x}}} \right] \cr
& {\text{Use the quotient rule}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {1 + x} \right)\frac{d}{{dx}}\left[ {{x^2}} \right] - {x^2}\frac{d}{{dx}}\left[ {1 + x} \right]}}{{{{\left( {1 + x} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {1 + x} \right)\left( {2x} \right) - {x^2}\left( 1 \right)}}{{{{\left( {1 + x} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{2{x^2} + 2x - {x^2}}}{{{{\left( {1 + x} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{{x^2} + 2x}}{{{{\left( {1 + x} \right)}^2}}} \cr
& {\text{Find }}m{\text{ at }}\left( {1,\frac{1}{2}} \right) \cr
& m = {\left. {\frac{{dy}}{{dx}}} \right|_{x = 1}} = \frac{{{{\left( 1 \right)}^2} + 2\left( 1 \right)}}{{{{\left( {1 + 1} \right)}^2}}} \cr
& m = \frac{3}{4} \cr
& \cr
& {\text{Use the Point}} - {\text{Slope Form of the Equation of a Line}} \cr
& y - {y_1} = m\left( {x - {x_1}} \right) \cr
& \underbrace {\left( {1,\frac{1}{2}} \right)}_{\left( {{x_1},{y_1}} \right)} \to x = 1{\text{ and }}{y_1} = \frac{1}{2} \cr
& {\text{Therefore}} \cr
& y - \frac{1}{2} = \frac{3}{4}\left( {x - 1} \right) \cr
& {\text{Simplify}} \cr
& y - \frac{1}{2} = \frac{3}{4}x - \frac{3}{4} \cr
& y = \frac{3}{4}x - \frac{3}{4} + \frac{1}{2} \cr
& y = \frac{3}{4}x - \frac{1}{4} \cr} $$
