Answer
$$\nabla w = \tan \left( 2 \right){\bf{i}} + 4{\sec ^2}\left( 2 \right){\bf{j}} + 4{\sec ^2}\left( 2 \right){\bf{k}}$$
Work Step by Step
$$\eqalign{
& w = x\tan \left( {y + z} \right),{\text{ }}\left( {4,3, - 1} \right) \cr
& {\text{Calculate the partial derivatives }}{w_x}{\text{, }}{w_y}{\text{ and }}{w_z} \cr
& {w_x} = \frac{\partial }{{\partial x}}\left[ {x\tan \left( {y + z} \right)} \right] \cr
& {w_x} = \tan \left( {y + z} \right) \cr
& \cr
& {w_y} = \frac{\partial }{{\partial y}}\left[ {x\tan \left( {y + z} \right)} \right] \cr
& {w_y} = x{\sec ^2}\left( {y + z} \right) \cr
& \cr
& {w_z} = \frac{\partial }{{\partial z}}\left[ {x\tan \left( {y + z} \right)} \right] \cr
& {w_z} = x{\sec ^2}\left( {y + z} \right) \cr
& \cr
& {\text{The gradient of }}w{\text{ is}} \cr
& \nabla w = {w_x}{\bf{i}} + {w_y}{\bf{j}} + {w_z}{\bf{k}} \cr
& \nabla w = \tan \left( {y + z} \right){\bf{i}} + x{\sec ^2}\left( {y + z} \right){\bf{j}} + x{\sec ^2}\left( {y + z} \right){\bf{k}} \cr
& {\text{At the point }}\left( {4,3, - 1} \right){\text{ the gradient is}} \cr
& \nabla w = \tan \left( {3 - 1} \right){\bf{i}} + 4{\sec ^2}\left( {3 - 1} \right){\bf{j}} + 4{\sec ^2}\left( {3 - 1} \right){\bf{k}} \cr
& \nabla w = \tan \left( 2 \right){\bf{i}} + 4{\sec ^2}\left( 2 \right){\bf{j}} + 4{\sec ^2}\left( 2 \right){\bf{k}} \cr} $$
