Answer
$$\eqalign{
& {z_s} = \left( {t + 1} \right){e^{st + s + t}} \cr
& {z_t} = \left( {s + 1} \right){e^{st + s + t}} \cr} $$
Work Step by Step
$$\eqalign{
& {\text{Let the functions }}z = {e^{x + y}},\,\,\,\,\,x = st{\text{ and }}y = s + t \cr
& \cr
& {\text{Calculate }}{z_s},{\text{ apply }}\frac{{\partial z}}{{\partial s}} = \frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial s}} + \frac{{\partial z}}{{\partial y}}\frac{{\partial y}}{{\partial s}} \cr
& \frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {{e^{x + y}}} \right] = {e^{x + y}} \cr
& \frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {{e^{x + y}}} \right] = {e^{x + y}} \cr
& \frac{{\partial x}}{{\partial s}} = \frac{\partial }{{\partial s}}\left[ {st} \right] = t \cr
& \frac{{\partial y}}{{\partial s}} = \frac{\partial }{{\partial s}}\left[ {s + t} \right] = 1 \cr
& {\text{Then}}{\text{,}} \cr
& \frac{{\partial z}}{{\partial s}} = {e^{x + y}}\left( t \right) + {e^{x + y}}\left( 1 \right) \cr
& \frac{{\partial z}}{{\partial s}} = t{e^{x + y}} + {e^{x + y}} \cr
& \frac{{\partial z}}{{\partial s}} = \left( {t + 1} \right){e^{x + y}} \cr
& {\text{Write in terms of }}s{\text{ and }}t \cr
& \frac{{\partial z}}{{\partial s}} = \left( {t + 1} \right){e^{st + s + t}} \cr
& \cr
& {\text{Calculate }}{z_t},{\text{ apply }}\frac{{\partial z}}{{\partial t}} = \frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial t}} + \frac{{\partial z}}{{\partial y}}\frac{{\partial y}}{{\partial t}} \cr
& \frac{{\partial x}}{{\partial t}} = \frac{\partial }{{\partial t}}\left[ {st} \right] = s \cr
& \frac{{\partial y}}{{\partial t}} = \frac{\partial }{{\partial t}}\left[ {s + t} \right] = 1 \cr
& {\text{Then}}{\text{,}} \cr
& \frac{{\partial z}}{{\partial t}} = {e^{x + y}}\left( s \right) + {e^{x + y}}\left( 1 \right) \cr
& \frac{{\partial z}}{{\partial t}} = s{e^{x + y}} + {e^{x + y}} \cr
& \frac{{\partial z}}{{\partial t}} = \left( {s + 1} \right){e^{x + y}} \cr
& {\text{Write in terms of }}s{\text{ and }}t \cr
& \frac{{\partial z}}{{\partial t}} = \left( {s + 1} \right){e^{st + s + t}} \cr} $$