Answer
$$\nabla f\left( {x,y} \right) = \frac{1}{{y + 1}}{\bf{i}} + \frac{{1 - x}}{{{{\left( {y + 1} \right)}^2}}}{\bf{j}},{\text{ }}\frac{{\sqrt 5 }}{4}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = \frac{{x + y}}{{y + 1}},{\text{ point }}\left( {0,1} \right) \cr
& {\text{Find the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\frac{x}{{y + 1}} + \frac{y}{{y + 1}}} \right] \cr
& {f_x}\left( {x,y} \right) = \frac{1}{{y + 1}} \cr
& and \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\frac{{x + y}}{{y + 1}}} \right] \cr
& {f_y}\left( {x,y} \right) = \frac{{\left( {y + 1} \right) - \left( {x + y} \right)}}{{{{\left( {y + 1} \right)}^2}}} \cr
& {f_y}\left( {x,y} \right) = \frac{{1 - x}}{{{{\left( {y + 1} \right)}^2}}} \cr
& {\text{Calculate }}\nabla f\left( {x,y} \right) \cr
& \nabla f\left( {x,y} \right) = {f_x}\left( {x,y} \right){\bf{i}} + {f_y}\left( {x,y} \right){\bf{j}} \cr
& \nabla f\left( {x,y} \right) = \frac{1}{{y + 1}}{\bf{i}} + \frac{{1 - x}}{{{{\left( {y + 1} \right)}^2}}}{\bf{j}} \cr
& {\text{At the given point the gradient is}} \cr
& \nabla f\left( {0,1} \right) = \frac{1}{{1 + 1}}{\bf{i}} + \frac{{1 - 0}}{{{{\left( {1 + 1} \right)}^2}}}{\bf{j}} \cr
& \nabla f\left( {0,1} \right) = \frac{1}{2}{\bf{i}} + \frac{1}{4}{\bf{j}} \cr
& {\text{The maximum value of }}{D_{\bf{u}}}f\left( {x,y} \right){\text{ is }}\left\| {\nabla f\left( {x,y} \right)} \right\| \cr
& \left\| {\nabla f\left( {0,1} \right)} \right\| = \left\| {\frac{1}{2}{\bf{i}} + \frac{1}{4}{\bf{j}}} \right\| \cr
& \left\| {\nabla f\left( {0,1} \right)} \right\| = \sqrt {\frac{1}{4} + \frac{1}{{16}}} \cr
& \left\| {\nabla f\left( {0,1} \right)} \right\| = \sqrt {\frac{5}{{16}}} \cr
& \left\| {\nabla f\left( {0,1} \right)} \right\| = \frac{{\sqrt 5 }}{4} \cr} $$
