Answer
$$0$$
Work Step by Step
$$\eqalign{
& g\left( {x,y} \right) = \ln \left( {4 + {x^2} + {y^2}} \right);\,\,\,\,\,\,P\left( { - 1,2} \right);\,\,\,\,\,\left\langle {2,1} \right\rangle \cr
& {\bf{u}} = \frac{{\left\langle {2,1} \right\rangle }}{{\sqrt {{{\left( 2 \right)}^2} + {{\left( 1 \right)}^2}} }} = \left\langle {\frac{2}{{\sqrt 5 }},\frac{1}{{\sqrt 5 }}} \right\rangle \cr
& {\bf{u}}{\text{ is a unit vector in the direction of }}\left\langle {2,1} \right\rangle \cr
& {\text{The gradient of }}g\left( {x,y} \right){\text{ is}} \cr
& {g_x}\left( {x,y} \right) = \frac{{2x}}{{4 + {x^2} + {y^2}}} \cr
& {g_y}\left( {x,y} \right) = \frac{{2y}}{{4 + {x^2} + {y^2}}} \cr
& \nabla g\left( {x,y} \right) = \frac{{2x}}{{4 + {x^2} + {y^2}}}{\bf{i}} + \frac{{2y}}{{4 + {x^2} + {y^2}}}{\bf{j}} \cr
& \nabla g\left( {x,y} \right) = \frac{2}{{4 + {x^2} + {y^2}}}\left( {x{\bf{i}} + y{\bf{j}}} \right) \cr
& \cr
& {\text{Calculate the gradient at the point }}P\left( { - 1,2} \right) \cr
& \nabla g\left( { - 1,2} \right) = \frac{2}{{4 + {{\left( { - 1} \right)}^2} + {{\left( 2 \right)}^2}}}\left( { - {\bf{i}} + 2{\bf{j}}} \right) \cr
& \nabla g\left( { - 1,2} \right) = \frac{2}{9}\left( { - {\bf{i}} + 2{\bf{j}}} \right) \cr
& \nabla g\left( { - 1,2} \right) = \left\langle { - \frac{2}{9},\frac{4}{9}} \right\rangle \cr
& \cr
& {\text{Computing the directional derivatives of }}h{\text{ at }}\left( { - 1,2} \right) \cr
& \operatorname{in} {\text{ the direction of the vector }}{\bf{u}} = \left\langle {\frac{2}{{\sqrt 5 }},\frac{1}{{\sqrt 5 }}} \right\rangle \cr
& {D_{\bf{u}}}g\left( {a,b} \right) = \nabla g\left( {a,b} \right) \cdot {\bf{u}} \cr
& {D_{\bf{u}}}g\left( { - 1,2} \right) = \left\langle { - \frac{2}{9},\frac{4}{9}} \right\rangle \cdot \left\langle {\frac{2}{{\sqrt 5 }},\frac{1}{{\sqrt 5 }}} \right\rangle \cr
& {D_{\bf{u}}}g\left( { - 1,2} \right) = - \frac{4}{{9\sqrt 5 }} + \frac{4}{{9\sqrt 5 }} \cr
& {D_{\bf{u}}}g\left( { - 1,2} \right) = 0 \cr} $$
