Answer
$$\eqalign{
& \left( {\bf{a}} \right)L\left( {x,y} \right) = \frac{2}{5}x - \frac{4}{5}y + \frac{3}{5} \cr
& \left( {\bf{b}} \right)L\left( {3.1, - 1.04} \right) = 4.96 \cr} $$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = \sqrt {{x^2} + {y^2}} ;\,\,\,\,\left( {3, - 4} \right)\,\,\,{\text{ and }}\,\,\,\left( {3.06, - 3.92} \right) \cr
& f\left( {3, - 4} \right) = \sqrt {{{\left( 3 \right)}^2} + {{\left( { - 4} \right)}^2}} \cr
& f\left( {3, - 4} \right) = 5 \cr
& {\text{The partial derivatives are}} \cr
& {f_x}\left( {x,y} \right) = \frac{x}{{\sqrt {{x^2} + {y^2}} }} \cr
& {f_y}\left( {x,y} \right) = \frac{y}{{\sqrt {{x^2} + {y^2}} }} \cr
& {\text{Evaluate the partial derivatives at }}\left( {3, - 4} \right) \cr
& {f_x}\left( {3, - 4} \right) = \frac{2}{{\sqrt {{{\left( 3 \right)}^2} + {{\left( { - 4} \right)}^2}} }} = \frac{2}{5} \cr
& {f_y}\left( {3, - 4} \right) = - \frac{4}{{\sqrt {{{\left( 3 \right)}^2} + {{\left( { - 4} \right)}^2}} }} = - \frac{4}{5} \cr
& \cr
& \left( a \right) \cr
& {\text{The linear approximation to the function at }}\,\left( {3, - 4,5} \right){\text{ is}} \cr
& L\left( {x,y} \right) = {f_x}\left( {a,b} \right)\left( {x - a} \right) + {f_y}\left( {a,b} \right)\left( {y - b} \right) + f\left( {a,b} \right) \cr
& L\left( {x,y} \right) = \frac{2}{5}\left( {x - 3} \right) - \frac{4}{5}\left( {y + 4} \right) + 5 \cr
& L\left( {x,y} \right) = \frac{2}{5}x - \frac{6}{5} - \frac{4}{5}y - \frac{{16}}{5} + 5 \cr
& L\left( {x,y} \right) = \frac{2}{5}x - \frac{4}{5}y + \frac{3}{5} \cr
& \cr
& \left( b \right){\text{Estimate }}\,\left( {3.06, - 3.92} \right) \cr
& L\left( {x,y} \right) = \frac{2}{5}\left( {3.06} \right) - \frac{4}{5}\left( { - 3.92} \right) + \frac{3}{5} \cr
& L\left( {3.1, - 1.04} \right) = 4.96 \cr} $$
