Answer
$$\eqalign{
& {f_x}\left( {x,y,z} \right) = y + z \cr
& {f_y}\left( {x,y,z} \right) = x + z \cr
& {f_z}\left( {x,y,z} \right) = x + y \cr} $$
Work Step by Step
$$\eqalign{
& f\left( {x,y,z} \right) = xy + xz + yz \cr
& {\text{Find the first partial derivative }}{f_x}\left( {x,y,z} \right) \cr
& {f_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {xy + xz + yz} \right] \cr
& {\text{treat }}y{\text{ and }}z{\text{ as a constant}}{\text{, then }} \cr
& {f_x}\left( {x,y,z} \right) = y\frac{\partial }{{\partial x}}\left[ x \right] + z\frac{\partial }{{\partial x}}\left[ x \right] + \frac{\partial }{{\partial x}}\left[ {yz} \right] \cr
& {f_x}\left( {x,y,z} \right) = y + z \cr
& \cr
& {\text{Find the first partial derivative }}{f_y}\left( {x,y,z} \right) \cr
& {f_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {xy + xz + yz} \right] \cr
& {\text{treat }}x{\text{ and }}z{\text{ as a constant}}{\text{, then }} \cr
& {f_y}\left( {x,y,z} \right) = x\frac{\partial }{{\partial y}}\left[ y \right] + \frac{\partial }{{\partial y}}\left[ {xz} \right] + z\frac{\partial }{{\partial y}}\left[ y \right] \cr
& {f_y}\left( {x,y,z} \right) = x + z \cr
& \cr
& {\text{Find the first partial derivative }}{f_z}\left( {x,y,z} \right) \cr
& {f_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {xy + xz + yz} \right] \cr
& {\text{treat }}x{\text{ and }}y{\text{ as a constant}}{\text{, then }} \cr
& {f_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {xy} \right] + x\frac{\partial }{{\partial z}}\left[ z \right] + y\frac{\partial }{{\partial z}}\left[ z \right] \cr
& {f_z}\left( {x,y,z} \right) = x + y \cr} $$
