Answer
${g_x}\left( {x,t} \right) = t{e^{xt}}{\text{ and }}{g_t}\left( {x,t} \right) = x{e^{xt}}$
Work Step by Step
$$\eqalign{
& g\left( {x,t} \right) = {e^{xt}} \cr
& \cr
& {\text{Differentiate with respect to }}x \cr
& {g_x}\left( {x,t} \right) = \frac{\partial }{{\partial x}}\left[ {{e^{xt}}} \right] \cr
& {\text{Using the Chain Rule}} \cr
& {g_x}\left( {x,t} \right) = {e^{xt}}\frac{\partial }{{\partial x}}\left[ {xt} \right] \cr
& {\text{Considering }}t{\text{ as a constant}} \cr
& {g_x}\left( {x,t} \right) = t{e^{xt}}\frac{\partial }{{\partial x}}\left[ x \right] \cr
& {g_x}\left( {x,t} \right) = t{e^{xt}}\left( 1 \right) \cr
& {g_x}\left( {x,t} \right) = t{e^{xt}} \cr
& \cr
& {\text{Differentiate with respect to }}t \cr
& {g_t}\left( {x,t} \right) = \frac{\partial }{{\partial t}}\left[ {{e^{xt}}} \right] \cr
& {\text{Using the Chain Rule}} \cr
& {g_t}\left( {x,t} \right) = {e^{xt}}\frac{\partial }{{\partial t}}\left[ {xt} \right] \cr
& {\text{Considering }}x{\text{ as a constant}} \cr
& {g_t}\left( {x,t} \right) = x{e^{xt}}\frac{\partial }{{\partial t}}\left[ t \right] \cr
& {g_t}\left( {x,t} \right) = x{e^{xt}}\left( 1 \right) \cr
& {g_t}\left( {x,t} \right) = x{e^{xt}}} $$