Answer
$$\eqalign{
& \left( a \right).{\text{ }}m = - 5 \cr
& \left( b \right).{\text{ }}y = - 5x + 3 \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{x + 3}}{{2x + 1}};{\text{ }}P\left( {0,3} \right) \cr
& \left( a \right){\text{ Calculate }}f'\left( x \right){\text{ using the definition of the derivative}} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{x + h + 3}}{{2\left( {x + h} \right) + 1}} - \frac{{x + 3}}{{2x + 1}}}}{h} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{2{x^2} + 2xh + h + 6x - 2x\left( {x + h} \right) - 6\left( {x + h} \right)}}{{\left[ {2\left( {x + h} \right) + 1} \right]\left( {2x + 1} \right)}}}}{h} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{ - 5h}}{{\left[ {2\left( {x + h} \right) + 1} \right]\left( {2x + 1} \right)}}}}{h} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{ - 5}}{{\left[ {2\left( {x + h} \right) + 1} \right]\left( {2x + 1} \right)}} \cr
& f'\left( x \right) = - \frac{5}{{{{\left( {2x + 1} \right)}^2}}} \cr
& {\text{Calculate the slope at }}P\left( {0,3} \right) \cr
& m = f'\left( 0 \right) = - \frac{5}{{{{\left( {2\left( 0 \right) + 1} \right)}^2}}} = - 5 \cr
& \cr
& \left( b \right){\text{ Find an equation of the tangent line at }}P\left( {0,3} \right) \cr
& y - 3 = - 5\left( {x - 0} \right) \cr
& y - 3 = - 5x \cr
& {\text{ }}y = - 5x + 3 \cr} $$
