Answer
$${\text{There are no critical numbers}}{\text{.}}$$
Work Step by Step
$$\eqalign{
& z = {e^{ - x}}\sin y \cr
& {\text{Let }}z = f\left( {x,y} \right) \cr
& f\left( {x,y} \right) = {e^{ - x}}\sin y \cr
& {\text{Calculate the first partial derivatives of }}f\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{e^{ - x}}\sin y} \right] \cr
& {f_x}\left( {x,y} \right) = \sin y\frac{\partial }{{\partial x}}\left[ {{e^{ - x}}} \right] \cr
& {f_x}\left( {x,y} \right) = \sin y\left( { - {e^{ - x}}} \right) \cr
& {f_x}\left( {x,y} \right) = - {e^{ - x}}\sin y \cr
& and \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{e^{ - x}}\sin y} \right] \cr
& {f_y}\left( {x,y} \right) = {e^{ - x}}\frac{\partial }{{\partial y}}\left[ {\sin y} \right] \cr
& {f_y}\left( {x,y} \right) = {e^{ - x}}\cos y \cr
& {\text{Setting both first partial derivatives equal to zero, we have}} \cr
& {f_x}\left( {x,y} \right) = 0,{\text{ }}{f_y}\left( {x,y} \right) = 0 \cr
& - {e^{ - x}}\sin y = 0,{\text{ }}{e^{ - x}}\cos y = 0 \cr
& {e^{ - x}}\sin y = 0,{\text{ }}{e^{ - x}}\cos y = 0 \cr
& {e^{ - x}}{\text{ is always positive, so:}}{\text{.}} \cr
& \sin y = 0,{\text{ }}\cos y = 0 \cr
& {\text{There are no real value }}y{\text{ at which both expressions}} \cr
& \sin y = 0{\text{ and }}\cos y = 0{\text{ }}\left( {{\text{graph below}}} \right){\text{ simultaneosly,}} \cr
& {\text{so there are no critical numbers}}{\text{.}} \cr} $$
