Answer
$${\text{ }}{\bf{a}}){\text{ }}m = 0,{\text{ }}{\bf{b}})y = 1$$
Work Step by Step
$$\eqalign{
& \ln \left( {x + y} \right) - x = 3{x^2} \cr
& {\text{We have that }}x = 0,{\text{ substituting}} \cr
& \ln \left( {0 + y} \right) - 0 = 3{\left( 0 \right)^2} \cr
& \ln y = 0 \cr
& y = 1 \cr
& {\text{For }}x = 0{\text{ we obtain the }} \cr
& {\text{Point }}\left( {0,1} \right) \cr
& \ln \left( {x + y} \right) - x = 3{x^2} \cr
& {\text{Differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left[ {\ln \left( {x + y} \right)} \right] - \frac{d}{{dx}}\left[ x \right] = \frac{d}{{dx}}\left[ {3{x^2}} \right] \cr
& {\text{Therefore,}} \cr
& \frac{1}{{x + y}}\frac{d}{{dx}}\left[ {x + y} \right] - \frac{d}{{dx}}\left[ x \right] = \frac{d}{{dx}}\left[ {3{x^2}} \right] \cr
& \frac{1}{{x + y}}\left( {1 + \frac{{dy}}{{dx}}} \right) - 1 = 6x \cr
& \frac{1}{{x + y}} + \frac{1}{{x + y}}\frac{{dy}}{{dx}} - 1 = 6x \cr
& \frac{1}{{x + y}}\frac{{dy}}{{dx}} = 6x + 1 - \frac{1}{{x + y}} \cr
& \frac{{dy}}{{dx}} = \left( {6x + 1} \right)\left( {x + y} \right) - 1 \cr
& \left( {\bf{a}} \right){\text{ Calculating the slope at the point}}\left( {0,1} \right) \cr
& m = {\left. {\frac{{dy}}{{dx}}} \right|_{\left( {10, - 1} \right)}} = \left( {0 + 1} \right)\left( {0 + 1} \right) - 1 \cr
& m = 0 \cr
& \cr
& \left( {\bf{b}} \right){\text{The equation of the tangent line at }}\left( {0, 1} \right){\text{ is}} \cr
& y - {y_1} = m\left( {x - {x_1}} \right) \cr
& y - 1 = 0\left( {x - 0} \right) \cr
& {\text{Simplify}} \cr
& y = 1 \cr} $$
