Answer
$$dz = - 3{e^{ - 3x}}\cos 6ydx - 6{e^{ - 3x}}\sin 6ydy$$
Work Step by Step
$$\eqalign{
& z = {e^{ - 3x}}\cos 6y \cr
& {\text{Let }}z = f\left( {x,y} \right):{\text{ }} \cr
& {\text{Calculate the partial derivative }}{f_x}\left( {x,y} \right){\text{, treating }}y{\text{ as a constant}} \cr
& {\text{ }}{f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{e^{ - 3x}}\cos 6y} \right] \cr
& {\text{ }}{f_x}\left( {x,y} \right) = \cos 6y\frac{\partial }{{\partial x}}\left[ {{e^{ - 3x}}} \right] \cr
& {\text{ }}{f_x}\left( {x,y} \right) = \cos 6y\left( { - 3{e^{ - 3x}}} \right) \cr
& {\text{ }}{f_x}\left( {x,y} \right) = - 3{e^{ - 3x}}\cos 6y \cr
& {\text{Calculate the partial derivative }}{f_y}\left( {x,y} \right){\text{, treating }}x{\text{ as a constant}} \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{e^{ - 3x}}\cos 6y} \right] \cr
& {\text{ }}{f_y}\left( {x,y} \right) = {e^{ - 3x}}\frac{\partial }{{\partial y}}\left[ {\cos 6y} \right] \cr
& {\text{ }}{f_y}\left( {x,y} \right) = {e^{ - 3x}}\left( { - 6\sin 6y} \right) \cr
& {\text{ }}{f_y}\left( {x,y} \right) = - 6{e^{ - 3x}}\sin 6y \cr
& \cr
& {\text{The total differential of }}z{\text{ is given by }}dz = {f_x}\left( {x,y} \right)dx + {f_y}\left( {x,y} \right)dy \cr
& {\text{Substitute the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& dz = - 3{e^{ - 3x}}\cos 6ydx - 6{e^{ - 3x}}\sin 6ydy \cr} $$
