Answer
$$\frac{1}{{\sqrt 5 }}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = x/\left( {x - y} \right);\,\,\,\,\,\,P\left( {4,1} \right);\,\,\,\,\,\left\langle { - 1,2} \right\rangle \cr
& {\bf{u}} = \frac{{\left\langle { - 1,2} \right\rangle }}{{\sqrt {{{\left( { - 1} \right)}^2} + {{\left( 2 \right)}^2}} }} = \left\langle { - \frac{1}{{\sqrt 5 }},\frac{2}{{\sqrt 5 }}} \right\rangle \cr
& {\bf{u}}{\text{ is a unit vector in the direction of }}\left\langle { - 1,2} \right\rangle \cr
& {\text{The gradient of }}f\left( {x,y} \right){\text{ is}} \cr
& {f_x}\left( {x,y} \right) = - \frac{y}{{{{\left( {x - y} \right)}^2}}} \cr
& {f_y}\left( {x,y} \right) = \frac{x}{{{{\left( {x - y} \right)}^2}}} \cr
& \nabla f\left( {x,y} \right) = - \frac{y}{{{{\left( {x - y} \right)}^2}}}{\bf{i}} + \frac{x}{{{{\left( {x - y} \right)}^2}}}{\bf{j}} \cr
& \nabla f\left( {x,y} \right) = \frac{1}{{{{\left( {x - y} \right)}^2}}}\left( { - y{\bf{i}} + x{\bf{j}}} \right) \cr
& \cr
& {\text{Calculate the gradient at the point }}P\left( {4,1} \right) \cr
& \nabla f\left( {4,1} \right) = \frac{1}{{{{\left( {4 - 1} \right)}^2}}}\left( { - {\bf{i}} + 4{\bf{j}}} \right) \cr
& \nabla f\left( {4,1} \right) = \frac{1}{9}\left( { - {\bf{i}} + 4{\bf{j}}} \right) \cr
& \nabla f\left( {4,1} \right) = \left\langle { - \frac{1}{9},\frac{4}{9}} \right\rangle \cr
& \cr
& {\text{Computing the directional derivatives of }}f{\text{ at }}\left( {4,1} \right) \cr
& \operatorname{in} {\text{ the direction of the vector }}{\bf{u}} = \left\langle { - \frac{3}{5},\frac{4}{5}} \right\rangle \cr
& {D_{\bf{u}}}f\left( {a,b} \right) = \nabla f\left( {a,b} \right) \cdot {\bf{u}} \cr
& {D_{\bf{u}}}f\left( {4,1} \right) = \left\langle { - \frac{1}{9},\frac{4}{9}} \right\rangle \cdot \left\langle { - \frac{1}{{\sqrt 5 }},\frac{2}{{\sqrt 5 }}} \right\rangle \cr
& {D_{\bf{u}}}f\left( {4,1} \right) = \frac{1}{{9\sqrt 5 }} + \frac{8}{{9\sqrt 5 }} \cr
& {D_{\bf{u}}}f\left( {4,1} \right) = \frac{1}{{\sqrt 5 }} \cr} $$
