Answer
$${f_{xyy}} = {f_{yxy}} = {f_{yyx}} = {z^2}{e^{ - x}}\sin yz$$
Work Step by Step
$$\eqalign{
& f\left( {x,y,z} \right) = {e^{ - x}}\sin yz \cr
& {\text{Find }}{f_{xyy}}\left( {x,y,z} \right) \cr
& {f_x} = \frac{\partial }{{\partial x}}\left[ {{e^{ - x}}\sin yz} \right] = - {e^{ - x}}\sin yz \cr
& {f_{xy}} = \frac{\partial }{{\partial y}}\left[ { - {e^{ - x}}\sin yz} \right] = - z{e^{ - x}}\cos yz \cr
& {f_{xyy}} = \frac{\partial }{{\partial y}}\left[ { - z{e^{ - x}}\cos yz} \right] = {z^2}{e^{ - x}}\sin yz \cr
& {\text{Find }}{f_{yxy}}\left( {x,y,z} \right) \cr
& {f_y} = \frac{\partial }{{\partial y}}\left[ {{e^{ - x}}\sin yz} \right] = z{e^{ - x}}\cos yz \cr
& {f_{yx}} = \frac{\partial }{{\partial x}}\left[ {z{e^{ - x}}\cos yz} \right] = - z{e^{ - x}}\cos yz \cr
& {f_{yxy}} = \frac{\partial }{{\partial y}}\left[ { - z{e^{ - x}}\cos yz} \right] = {z^2}{e^{ - x}}\sin yz \cr
& {\text{Find }}{f_{yyx}}\left( {x,y,z} \right) \cr
& {f_{yy}} = \frac{\partial }{{\partial y}}\left[ {z{e^{ - x}}\cos yz} \right] = - {z^2}{e^{ - x}}\sin yz \cr
& {f_{yyx}} = \frac{\partial }{{\partial x}}\left[ { - {z^2}{e^{ - x}}\sin yz} \right] = {z^2}{e^{ - x}}\sin yz \cr
& {\text{Therefore,}} \cr
& {f_{xyy}} = {f_{yxy}} = {f_{yyx}} = {z^2}{e^{ - x}}\sin yz \cr} $$
