Answer
$${h_x}\left( {x,y,z} \right) = {h_y}\left( {x,y,z} \right) = {h_z}\left( {x,y,z} \right) = - \sin \left( {x + y + z} \right)$$
Work Step by Step
$$\eqalign{
& h\left( {x,y,z} \right) = \cos \left( {x + y + z} \right) \cr
& {\text{Find the first partial derivative }}{h_x}\left( {x,y,z} \right) \cr
& {h_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {\cos \left( {x + y + z} \right)} \right] \cr
& {\text{treat }}y{\text{ and }}z{\text{ as a constant}}{\text{, then }} \cr
& {h_x}\left( {x,y,z} \right) = - \sin \left( {x + y + z} \right)\frac{\partial }{{\partial x}}\left[ {x + y + z} \right] \cr
& {h_x}\left( {x,y,z} \right) = - \sin \left( {x + y + z} \right)\left( 1 \right) \cr
& {h_x}\left( {x,y,z} \right) = - \sin \left( {x + y + z} \right) \cr
& \cr
& {\text{Find the first partial derivative }}{h_y}\left( {x,y,z} \right) \cr
& {h_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {\cos \left( {x + y + z} \right)} \right] \cr
& {\text{treat }}x{\text{ and }}z{\text{ as a constant}}{\text{, then }} \cr
& {h_y}\left( {x,y,z} \right) = - \sin \left( {x + y + z} \right)\frac{\partial }{{\partial y}}\left[ {x + y + z} \right] \cr
& {h_y}\left( {x,y,z} \right) = - \sin \left( {x + y + z} \right)\left( 1 \right) \cr
& {h_y}\left( {x,y,z} \right) = - \sin \left( {x + y + z} \right) \cr
& \cr
& {\text{Find the first partial derivative }}{h_z}\left( {x,y,z} \right) \cr
& {h_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {\cos \left( {x + y + z} \right)} \right] \cr
& {\text{treat }}x{\text{ and }}y{\text{ as a constant}}{\text{, then }} \cr
& {h_z}\left( {x,y,z} \right) = - \sin \left( {x + y + z} \right)\frac{\partial }{{\partial y}}\left[ {x + y + z} \right] \cr
& {h_z}\left( {x,y,z} \right) = - \sin \left( {x + y + z} \right)\left( 1 \right) \cr
& {h_z}\left( {x,y,z} \right) = - \sin \left( {x + y + z} \right) \cr} $$
