Answer
$$\eqalign{
& \left( {\bf{a}} \right)L\left( {x,y} \right) = x + y \cr
& \left( {\bf{b}} \right)L\left( {0.1, - 0.2} \right) = - 0.1 \cr} $$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = \ln \left( {1 + x + y} \right);\,\,\,\,\left( {0,0} \right)\,\,\,{\text{ and }}\,\,\,\left( {0.1, - 0.2} \right) \cr
& f\left( {0,0} \right) = \ln \left( {1 + 0 + 0} \right) \cr
& f\left( {0,0} \right) = 0 \cr
& {\text{The partial derivatives are}} \cr
& {f_x}\left( {x,y} \right) = \frac{1}{{1 + x + y}} \cr
& {f_y}\left( {x,y} \right) = \frac{1}{{1 + x + y}} \cr
& {\text{Evaluate the partial derivatives at }}\left( {0,0} \right) \cr
& {f_x}\left( {0,0} \right) = 1 \cr
& {f_y}\left( {0,0} \right) = 1 \cr
& \cr
& \left( a \right) \cr
& {\text{The linear approximation to the function at }}\,\left( {0,0,0} \right){\text{ is}} \cr
& L\left( {x,y} \right) = {f_x}\left( {a,b} \right)\left( {x - a} \right) + {f_y}\left( {a,b} \right)\left( {y - b} \right) + f\left( {a,b} \right) \cr
& L\left( {x,y} \right) = \left( {x - 0} \right) + \left( {y + 0} \right) + 0 \cr
& L\left( {x,y} \right) = x + y \cr
& \cr
& \left( b \right){\text{Estimate }}\,\left( {0.1, - 0.2} \right) \cr
& L\left( {0.1, - 0.2} \right) = 0.1 - 0.2 \cr
& L\left( {0.1, - 0.2} \right) = - 0.1 \cr} $$