Answer
$$\eqalign{
& {h_x}\left( {x,y,z} \right) = z{\left( {1 + x + 2y} \right)^{z - 1}} \cr
& {h_y}\left( {x,y,z} \right) = 2z{\left( {1 + x + 2y} \right)^{z - 1}} \cr
& {h_z}\left( {x,y,z} \right) = {\left( {1 + x + 2y} \right)^z}\ln \left( {1 + x + 2y} \right) \cr} $$
Work Step by Step
$$\eqalign{
& h\left( {x,y,z} \right) = {\left( {1 + x + 2y} \right)^z} \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[ {{{\left( {1 + x + 2y} \right)}^z}} \right] \cr
& {\text{treat }}y{\text{ and }}z{\text{ as a constant}}{\text{, then use power rule }}\left( {{u^n}} \right)' = n{u^{n - 1}}\left( {u'} \right){\text{ }} \cr
& {h_x}\left( {x,y,z} \right) = z{\left( {1 + x + 2y} \right)^{z - 1}}\frac{\partial }{{\partial x}}\left[ {1 + x + 2y} \right] \cr
& {h_x}\left( {x,y,z} \right) = z{\left( {1 + x + 2y} \right)^{z - 1}}\left( 1 \right) \cr
& {h_x}\left( {x,y,z} \right) = z{\left( {1 + x + 2y} \right)^{z - 1}} \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[ {{{\left( {1 + x + 2y} \right)}^z}} \right] \cr
& {\text{treat }}x{\text{ and }}z{\text{ as a constant}}{\text{, then use power rule }}\left( {{u^n}} \right)' = n{u^{n - 1}}\left( {u'} \right){\text{ }} \cr
& {h_y}\left( {x,y,z} \right) = z{\left( {1 + x + 2y} \right)^{z - 1}}\frac{\partial }{{\partial y}}\left[ {1 + x + 2y} \right] \cr
& {h_y}\left( {x,y,z} \right) = z{\left( {1 + x + 2y} \right)^{z - 1}}\left( 2 \right) \cr
& {h_y}\left( {x,y,z} \right) = 2z{\left( {1 + x + 2y} \right)^{z - 1}} \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[ {{{\left( {1 + x + 2y} \right)}^z}} \right] \cr
& {\text{treat }}x{\text{ and }}y{\text{ as a constant}}{\text{, then use the rule }}\left( {{a^u}} \right)' = {a^u}\ln a\left( {u'} \right){\text{ }} \cr
& {h_z}\left( {x,y,z} \right) = {\left( {1 + x + 2y} \right)^z}\ln \left( {1 + x + 2y} \right)\frac{\partial }{{\partial z}}\left[ z \right] \cr
& {h_z}\left( {x,y,z} \right) = {\left( {1 + x + 2y} \right)^z}\ln \left( {1 + x + 2y} \right)\frac{\partial }{{\partial z}}\left[ z \right] \cr
& {h_z}\left( {x,y,z} \right) = {\left( {1 + x + 2y} \right)^z}\ln \left( {1 + x + 2y} \right) \cr} $$
