Answer
$${f_{xyy}} = {f_{yxy}} = {f_{yyx}} = 0$$
Work Step by Step
$$\eqalign{
& f\left( {x,y,z} \right) = {x^2} - 3xy + 4yz + {z^3} \cr
& {\text{Find }}{f_{xyy}}\left( {x,y,z} \right) \cr
& {f_x} = \frac{\partial }{{\partial x}}\left[ {{x^2} - 3xy + 4yz + {z^3}} \right] = 2x - 3y \cr
& {f_{xy}} = \frac{\partial }{{\partial y}}\left[ {2x - 3y} \right] = - 3 \cr
& {f_{xyy}} = \frac{\partial }{{\partial y}}\left[ { - 3} \right] = 0 \cr
& {\text{Find }}{f_{yxy}}\left( {x,y,z} \right) \cr
& {f_y} = \frac{\partial }{{\partial y}}\left[ {{x^2} - 3xy + 4yz + {z^3}} \right] = 3x + 4z \cr
& {f_{yx}} = \frac{\partial }{{\partial x}}\left[ {3x + 4z} \right] = 3 \cr
& {f_{yxy}} = \frac{\partial }{{\partial y}}\left[ 3 \right] = 0 \cr
& {\text{Find }}{f_{yyx}}\left( {x,y,z} \right) \cr
& {f_{yy}} = \frac{\partial }{{\partial y}}\left[ {3x + 4z} \right] = 0 \cr
& {f_{yyx}} = \frac{\partial }{{\partial x}}\left[ 0 \right] = 0 \cr
& {\text{Therefore,}} \cr
& {f_{xyy}} = {f_{yxy}} = {f_{yyx}} = 0 \cr} $$
