Answer
$$\eqalign{
& {z_s} = \left( {2 - \sin t} \right)\sin s \cr
& {z_t} = \left( {\cos s + 3} \right)\cos t \cr} $$
Work Step by Step
$$\eqalign{
& {\text{Let the functions }}z = xy - 2x + 3y,\,\,\,\,\,x = \cos s{\text{ and }}y = \sin t \cr
& \cr
& {\text{Calculate }}{z_s},{\text{ apply }}\frac{{\partial z}}{{\partial s}} = \frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial s}} + \frac{{\partial z}}{{\partial y}}\frac{{\partial y}}{{\partial s}} \cr
& \frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {xy - 2x + 3y} \right] = y - 2 \cr
& \frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {xy - 2x + 3y} \right] = x + 3 \cr
& \frac{{\partial x}}{{\partial s}} = \frac{\partial }{{\partial s}}\left[ {\cos s} \right] = - \sin s \cr
& \frac{{\partial y}}{{\partial s}} = \frac{\partial }{{\partial s}}\left[ {\sin t} \right] = 0 \cr
& {\text{Then}}{\text{,}} \cr
& \frac{{\partial z}}{{\partial s}} = \left( {y - 2} \right)\left( { - \sin s} \right) + \left( {x + 3} \right)\left( 0 \right) \cr
& \frac{{\partial z}}{{\partial s}} = \left( {2 - y} \right)\sin s \cr
& {\text{Write in terms of }}s{\text{ and }}t \cr
& \frac{{\partial z}}{{\partial s}} = \left( {2 - \sin t} \right)\sin s \cr
& \cr
& {\text{Calculate }}{z_t},{\text{ apply }}\frac{{\partial z}}{{\partial t}} = \frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial t}} + \frac{{\partial z}}{{\partial y}}\frac{{\partial y}}{{\partial t}} \cr
& \frac{{\partial x}}{{\partial t}} = \frac{\partial }{{\partial t}}\left[ {\cos s} \right] = 0 \cr
& \frac{{\partial y}}{{\partial t}} = \frac{\partial }{{\partial t}}\left[ {\sin t} \right] = \cos t \cr
& {\text{Then}}{\text{,}} \cr
& \frac{{\partial z}}{{\partial s}} = \left( {y - 2} \right)\left( 0 \right) + \left( {x + 3} \right)\left( {\cos t} \right) \cr
& \frac{{\partial z}}{{\partial t}} = \left( {x + 3} \right)\cos t \cr
& {\text{Write in terms of }}s{\text{ and }}t \cr
& \frac{{\partial z}}{{\partial t}} = \left( {\cos s + 3} \right)\cos t \cr} $$
