Answer
$$\left\| {\nabla f\left( {3,2} \right)} \right\| = \frac{{\sqrt {13} }}{6}$$
Work Step by Step
$$\eqalign{
& \nabla f\left( {x,y} \right) = - \frac{1}{3}{\bf{i}} - \frac{1}{2}{\bf{j}} \cr
& {\text{The gradient at }}\left( {3,2} \right){\text{ is}} \cr
& \nabla f\left( {3,2} \right) = - \frac{1}{3}{\bf{i}} - \frac{1}{2}{\bf{j}} \cr
& {\text{The maximum value of }}{D_{\bf{u}}}f\left( {x,y} \right){\text{ is }}\left\| {\nabla f\left( {x,y} \right)} \right\| \cr
& \left\| {\nabla f\left( {3,2} \right)} \right\| = \left\| { - \frac{1}{3}{\bf{i}} - \frac{1}{2}{\bf{j}}} \right\| \cr
& \left\| {\nabla f\left( {3,2} \right)} \right\| = \sqrt {\frac{1}{9} + \frac{1}{4}} \cr
& \left\| {\nabla f\left( {3,2} \right)} \right\| = \sqrt {\frac{{13}}{{36}}} \cr
& \left\| {\nabla f\left( {3,2} \right)} \right\| = \frac{{\sqrt {13} }}{6} \cr} $$
