Answer
$1$
Work Step by Step
$$\eqalign{
& f\left( {x,y,z} \right) = {x^2}y + {y^2}z,{\text{ }}\left( {1,2,3} \right),{\text{ }}{\bf{v}} = \left\langle {2, - 1,2} \right\rangle \cr
& {\text{Calculate the partial derivatives }} \cr
& {f_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {{x^2}y + {y^2}z} \right] \cr
& {f_x}\left( {x,y,z} \right) = 2xy \cr
& {f_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {{x^2}y + {y^2}z} \right] \cr
& {f_y}\left( {x,y,z} \right) = {x^2} + 2yz \cr
& and \cr
& {f_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {{x^2}y + {y^2}z} \right] \cr
& {f_z}\left( {x,y,z} \right) = {y^2} \cr
& \cr
& {\text{The gradient of the function is given by }} \cr
& \nabla f\left( {x,y,z} \right) = {f_x}\left( {x,y,z} \right){\bf{i}} + {f_y}\left( {x,y,z} \right){\bf{j}} + {f_z}\left( {x,y,z} \right){\bf{k}} \cr
& {\text{Substituting the partial derivatives}}{\text{, we obtain}} \cr
& \nabla f\left( {x,y,z} \right) = 2xy{\bf{i}} + \left( {{x^2} + 2yz} \right){\bf{j}} + {y^2}{\bf{k}} \cr
& {\text{Calculate the gradient at the point }}\left( {1,2,3} \right) \cr
& \nabla f\left( {1,2,3} \right) = 2\left( 1 \right)\left( 2 \right){\bf{i}} + \left( {{1^2} + 2\left( 2 \right)\left( 3 \right)} \right){\bf{j}} + {\left( 2 \right)^2}{\bf{k}} \cr
& \nabla f\left( {1,2,3} \right) = 4{\bf{i}} + 13{\bf{j}} + 4{\bf{k}} \cr
& \cr
& {\text{The directional derivative of }}f\left( {x,y,z} \right){\text{ at }}\left( {{x_0},{y_0},{z_0}} \right){\text{ in the}} \cr
& {\text{direction of a unit vector }}{\bf{u}}{\text{ is}} \cr
& {D_{\bf{u}}}f\left( {x,y,z} \right) = \nabla f\left( {x,y,z} \right) \cdot {\bf{u}} \cr
& {\text{Express }}{\bf{v}} = \left\langle {2, - 1,2} \right\rangle {\text{ as a unit vector }}{\bf{u}} \cr
& {\bf{u}} = \frac{{\bf{v}}}{{\left| {\bf{v}} \right|}} = \frac{{\left\langle {2, - 1,2} \right\rangle }}{{\left| {\left\langle {2, - 1,2} \right\rangle } \right|}} = \frac{{\left\langle {2, - 1,2} \right\rangle }}{{\sqrt {4 + 1 + 4} }} = \left\langle {\frac{2}{3}, - \frac{1}{3},\frac{2}{3}} \right\rangle \cr
& {\text{Therefore at the point }}\left( {1,2,3} \right){\text{ }} \cr
& {D_{\bf{u}}}f\left( {1,2,3} \right) = \nabla f\left( {1,2,3} \right) \cdot {\bf{u}} \cr
& {D_{\bf{u}}}f\left( {1,2,3} \right) = \left( {4{\bf{i}} + 13{\bf{j}} + 4{\bf{k}}} \right) \cdot \left( {\frac{2}{3}{\bf{i}} - \frac{1}{3}{\bf{j}} + \frac{2}{3}{\bf{k}}} \right) \cr
& {D_{\bf{u}}}f\left( {1,2,3} \right) = \left( 4 \right)\left( {\frac{2}{3}} \right) + \left( {13} \right)\left( { - \frac{1}{3}} \right) + \left( 4 \right)\left( {\frac{2}{3}} \right) \cr
& {\text{Simplifying}} \cr
& {D_{\bf{u}}}f\left( {1,2,3} \right) = \frac{8}{3} - \frac{{13}}{3} + \frac{8}{3} \cr
& {D_{\bf{u}}}f\left( {1,2,3} \right) = 1 \cr} $$
