Answer
$$\eqalign{
& {r_{xx}} = - \frac{1}{{{{\left( {x + y} \right)}^2}}},\,\,\,\,\,\,{r_{yy}} = - \frac{1}{{{{\left( {x + y} \right)}^2}}} \cr
& {r_{xy}} = - \frac{1}{{{{\left( {x + y} \right)}^2}}},\,\,\,\,\,\,{r_{yx}} = - \frac{1}{{{{\left( {x + y} \right)}^2}}} \cr} $$
Work Step by Step
$$\eqalign{
& r = \ln \left| {x + y} \right| \cr
& {\text{Find }}{r_x}{\text{ and }}{r_y} \cr
& {r_x} = \frac{\partial }{{\partial x}}\left[ {\ln \left| {x + y} \right|} \right] \cr
& {\text{treat }}y{\text{ as a constant and use }}\left( {\ln u} \right)' = \frac{1}{u}\left( {u'} \right) \cr
& {r_x} = \frac{1}{{x + y}}\frac{\partial }{{\partial x}}\left[ {x + y} \right] \cr
& {\text{then}} \cr
& {r_x} = \frac{1}{{x + y}}\left( 1 \right) \cr
& {r_x} = \frac{1}{{x + y}} \cr
& \cr
& {r_y} = \frac{\partial }{{\partial y}}\left[ {\ln \left| {x + y} \right|} \right] \cr
& {\text{treat }}x{\text{ as a constant and use }}\left( {\ln u} \right)' = \frac{1}{u}\left( {u'} \right) \cr
& {r_y} = \frac{1}{{x + y}}\frac{\partial }{{\partial y}}\left[ {x + y} \right] \cr
& {\text{then}} \cr
& {r_y} = \frac{1}{{x + y}}\left( 1 \right) \cr
& {r_y} = \frac{1}{{x + y}} \cr
& \cr
& {\text{find the second - order partial derivatives}} \cr
& {r_{xy}} = \frac{\partial }{{\partial y}}\left[ {\frac{1}{{x + y}}} \right] \cr
& {r_{xy}} = \frac{\partial }{{\partial y}}\left[ {{{\left( {x + y} \right)}^{ - 1}}} \right] \cr
& {r_{xy}} = - {\left( {x + y} \right)^{ - 2}}\frac{\partial }{{\partial y}}\left[ {x + y} \right] \cr
& {r_{xy}} = - \frac{1}{{{{\left( {x + y} \right)}^2}}} \cr
& and \cr
& {r_{yx}} = \frac{\partial }{{\partial x}}\left[ {\frac{1}{{x + y}}} \right] \cr
& {r_{yx}} = \frac{\partial }{{\partial x}}\left[ {{{\left( {x + y} \right)}^{ - 1}}} \right] \cr
& {r_{yx}} = - {\left( {x + y} \right)^{ - 2}}\frac{\partial }{{\partial x}}\left[ {x + y} \right] \cr
& {r_{yx}} = - \frac{1}{{{{\left( {x + y} \right)}^2}}} \cr
& \cr
& {r_{xx}} = \frac{\partial }{{\partial x}}\left( {\frac{1}{{x + y}}} \right) \cr
& {r_{xx}} = - \frac{1}{{{{\left( {x + y} \right)}^2}}} \cr
& and \cr
& {r_{yy}} = \frac{\partial }{{\partial x}}\left( {\frac{1}{{x + y}}} \right) \cr
& {r_{yy}} = - \frac{1}{{{{\left( {x + y} \right)}^2}}} \cr} $$