Answer
$$\eqalign{
& \nabla h\left( {x,y} \right) = - y\sin \left( {x - y} \right){\bf{i}} + \left[ {\cos \left( {x - y} \right) + y\sin \left( {x - y} \right)} \right]{\bf{j}} \cr
& \left\| {\nabla h\left( {0,\frac{\pi }{3}} \right)} \right\| = \frac{{\sqrt {2{\pi ^2} + 3 - 2\sqrt 3 \pi } }}{{\sqrt {12} }} \cr} $$
Work Step by Step
$$\eqalign{
& h\left( {x,y} \right) = y\cos \left( {x - y} \right),{\text{ point }}\left( {0,\frac{\pi }{3}} \right) \cr
& {\text{Find the partial derivatives }}{h_x}\left( {x,y} \right){\text{ and }}{h_y}\left( {x,y} \right) \cr
& {h_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {y\cos \left( {x - y} \right)} \right] \cr
& {h_x}\left( {x,y} \right) = - y\sin \left( {x - y} \right) \cr
& and \cr
& {h_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {y\cos \left( {x - y} \right)} \right] \cr
& {h_y}\left( {x,y} \right) = \cos \left( {x - y} \right) + y\sin \left( {x - y} \right) \cr
& {\text{Calculate }}\nabla h\left( {x,y} \right) \cr
& \nabla h\left( {x,y} \right) = - y\sin \left( {x - y} \right){\bf{i}} + \left[ {\cos \left( {x - y} \right) + y\sin \left( {x - y} \right)} \right]{\bf{j}} \cr
& {\text{At the given point the gradient is}} \cr
& \nabla h\left( {0,\frac{\pi }{3}} \right) = - \frac{\pi }{3}\sin \left( { - \frac{\pi }{3}} \right){\bf{i}} + \left[ {\cos \left( { - \frac{\pi }{3}} \right) + \frac{\pi }{3}\sin \left( { - \frac{\pi }{3}} \right)} \right]{\bf{j}} \cr
& \nabla h\left( {0,\frac{\pi }{3}} \right) = \frac{{\sqrt 3 \pi }}{6}{\bf{i}} + \left( {\frac{1}{2} - \frac{{\sqrt 3 }}{6}\pi } \right){\bf{j}} \cr
& {\text{The maximum value of }}{D_{\bf{u}}}h\left( {x,y} \right){\text{ is }}\left\| {\nabla h\left( {x,y} \right)} \right\| \cr
& \left\| {\nabla h\left( {0,\frac{\pi }{3}} \right)} \right\| = \left\| {\frac{{\sqrt 3 }}{6}\pi {\bf{i}} + \left( {\frac{1}{2} - \frac{{\sqrt 3 }}{6}\pi } \right){\bf{j}}} \right\| \cr
& \left\| {\nabla h\left( {0,\frac{\pi }{3}} \right)} \right\| = \sqrt {\frac{3}{{36}}{\pi ^2} + \frac{1}{4} - \frac{{\sqrt 3 }}{6}\pi + \frac{3}{{36}}{\pi ^2}} \cr
& \left\| {\nabla h\left( {0,\frac{\pi }{3}} \right)} \right\| = \sqrt {\frac{1}{6}{\pi ^2} + \frac{1}{4} - \frac{{\sqrt 3 }}{6}\pi } \cr
& \left\| {\nabla h\left( {0,\frac{\pi }{3}} \right)} \right\| = \sqrt {\frac{{2{\pi ^2} + 3 - 2\sqrt 3 \pi }}{{12}}} \cr
& \left\| {\nabla h\left( {0,\frac{\pi }{3}} \right)} \right\| = \frac{{\sqrt {2{\pi ^2} + 3 - 2\sqrt 3 \pi } }}{{\sqrt {12} }} \cr} $$
