Answer
$$\nabla g\left( {x,y} \right) = \frac{2}{{3\left( {{x^2} + {y^2}} \right)}}\left( {x{\bf{i}} + y{\bf{j}}} \right),{\text{ }}\frac{1}{{15}}\sqrt {20} $$
Work Step by Step
$$\eqalign{
& g\left( {x,y} \right) = \ln \root 3 \of {{x^2} + {y^2}} ,{\text{ point }}\left( {1,2} \right) \cr
& g\left( {x,y} \right) = \ln {\left( {{x^2} + {y^2}} \right)^{1/3}} \cr
& g\left( {x,y} \right) = \frac{1}{3}\ln \left( {{x^2} + {y^2}} \right) \cr
& {\text{Find the partial derivatives }}{g_x}\left( {x,y} \right){\text{ and }}{g_y}\left( {x,y} \right) \cr
& {g_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\frac{1}{3}\ln \left( {{x^2} + {y^2}} \right)} \right] \cr
& {g_x}\left( {x,y} \right) = \frac{1}{3}\left( {\frac{{2x}}{{{x^2} + {y^2}}}} \right) \cr
& {g_x}\left( {x,y} \right) = \frac{2}{3}\left( {\frac{x}{{{x^2} + {y^2}}}} \right) \cr
& and \cr
& {g_y}\left( {x,y} \right) = \frac{2}{3}\left( {\frac{y}{{{x^2} + {y^2}}}} \right) \cr
& {\text{Calculate }}\nabla g\left( {x,y} \right) \cr
& \nabla g\left( {x,y} \right) = {g_x}\left( {x,y} \right){\bf{i}} + {g_y}\left( {x,y} \right){\bf{j}} \cr
& \nabla g\left( {x,y} \right) = \frac{2}{3}\left( {\frac{x}{{{x^2} + {y^2}}}} \right){\bf{i}} + \frac{2}{3}\left( {\frac{y}{{{x^2} + {y^2}}}} \right){\bf{j}} \cr
& \nabla g\left( {x,y} \right) = \frac{2}{{3\left( {{x^2} + {y^2}} \right)}}\left( {x{\bf{i}} + y{\bf{j}}} \right) \cr
& \cr
& {\text{Evaluate }}\nabla g\left( {1,2} \right) \cr
& \nabla g\left( {1,2} \right) = \frac{2}{{3\left( {{{\left( 1 \right)}^2} + {{\left( 2 \right)}^2}} \right)}}\left( {{\bf{i}} + 2{\bf{j}}} \right) \cr
& \nabla g\left( {1,2} \right) = \frac{2}{{15}}\left( {{\bf{i}} + 2{\bf{j}}} \right) \cr
& \cr
& {\text{The maximum value of }}{D_{\bf{u}}}g\left( {x,y} \right){\text{ is }}\left\| {\nabla g\left( {x,y} \right)} \right\| \cr
& \left\| {\nabla g\left( {0,5} \right)} \right\| = \left\| {\frac{2}{{15}}\left( {{\bf{i}} + 2{\bf{j}}} \right)} \right\| \cr
& \left\| {\nabla g\left( {0,5} \right)} \right\| = \sqrt {\frac{4}{{{{15}^2}}} + \frac{{16}}{{{{15}^2}}}} \cr
& \left\| {\nabla g\left( {0,5} \right)} \right\| = \frac{1}{{15}}\sqrt {20} \cr} $$
