Answer
$${f_{xyy}} = {f_{yxy}} = {f_{yyx}} = - \frac{{12z}}{{{{\left( {x + y} \right)}^4}}}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y,z} \right) = \frac{{2z}}{{x + y}} \cr
& {\text{Find }}{f_{xyy}}\left( {x,y,z} \right) \cr
& {f_x} = \frac{\partial }{{\partial x}}\left[ {\frac{{2z}}{{x + y}}} \right] = - \frac{{2z}}{{{{\left( {x + y} \right)}^2}}} \cr
& {f_{xy}} = \frac{\partial }{{\partial y}}\left[ { - \frac{{2z}}{{{{\left( {x + y} \right)}^2}}}} \right] = - \frac{\partial }{{\partial y}}\left[ { - 2z{{\left( {x + y} \right)}^{ - 2}}} \right] = 4z{\left( {x + y} \right)^{ - 3}} \cr
& {f_{xyy}} = \frac{\partial }{{\partial y}}\left[ {4z{{\left( {x + y} \right)}^{ - 3}}} \right] = - 12z{\left( {x + y} \right)^{ - 4}} = - \frac{{12z}}{{{{\left( {x + y} \right)}^4}}} \cr
& {\text{Find }}{f_{yxy}}\left( {x,y,z} \right) \cr
& {f_y} = \frac{\partial }{{\partial y}}\left[ {\frac{{2z}}{{x + y}}} \right] = - \frac{{2z}}{{{{\left( {x + y} \right)}^2}}} \cr
& {f_{yx}} = \frac{\partial }{{\partial x}}\left[ { - 2z{{\left( {x + y} \right)}^{ - 2}}} \right] = 4z{\left( {x + y} \right)^{ - 3}} \cr
& {f_{yxy}} = \frac{\partial }{{\partial y}}\left[ {4z{{\left( {x + y} \right)}^{ - 3}}} \right] = - \frac{{12z}}{{{{\left( {x + y} \right)}^4}}} \cr
& {\text{Find }}{f_{yyx}}\left( {x,y,z} \right) \cr
& {f_{yy}} = \frac{\partial }{{\partial y}}\left[ { - 2z{{\left( {x + y} \right)}^{ - 2}}} \right] = 4z{\left( {x + y} \right)^{ - 3}} \cr
& {f_{yyx}} = \frac{\partial }{{\partial x}}\left[ {4z{{\left( {x + y} \right)}^{ - 3}}} \right] = - \frac{{12z}}{{{{\left( {x + y} \right)}^4}}} \cr
& {\text{Therefore,}} \cr
& {f_{xyy}} = {f_{yxy}} = {f_{yyx}} = - \frac{{12z}}{{{{\left( {x + y} \right)}^4}}} \cr} $$
