Answer
$$\eqalign{
& {p_{uu}}\left( {u,v} \right) = \frac{{2{v^2} + 8 - 2{u^2}}}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}},\,\,\,\,\,\,\,\,{\text{ }}{p_{vv}}\left( {u,v} \right) = \frac{{2{u^2} + 8 - 2{v^2}}}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}} \cr
& {p_{uv}}\left( {u,v} \right) = {p_{vu}}\left( {u,v} \right) = - \frac{{4uv}}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}} \cr} $$
Work Step by Step
$$\eqalign{
& p\left( {u,v} \right) = \ln \left( {{u^2} + {v^2} + 4} \right) \cr
& {\text{Find the first partial derivatives }}{p_u}\left( {u,v} \right){\text{ and }}{p_v}\left( {u,v} \right){\text{ then}} \cr
& {p_u}\left( {u,v} \right) = \frac{\partial }{{\partial u}}\left[ {\ln \left( {{u^2} + {v^2} + 4} \right)} \right] \cr
& {\text{treat }}v{\text{ as a constant}}{\text{, then}} \cr
& {p_u}\left( {u,v} \right) = \frac{1}{{{u^2} + {v^2} + 4}}\frac{\partial }{{\partial u}}\left[ {{u^2} + {v^2} + 4} \right] \cr
& {p_u}\left( {u,v} \right) = \frac{1}{{{u^2} + {v^2} + 4}}\left( {2u} \right) \cr
& {p_u}\left( {u,v} \right) = \frac{{2u}}{{{u^2} + {v^2} + 4}} \cr
& and \cr
& {p_v}\left( {u,v} \right) = \frac{\partial }{{\partial v}}\left[ {\ln \left( {{u^2} + {v^2} + 4} \right)} \right] \cr
& {\text{treat }}x{\text{ as a constant}}{\text{, then }} \cr
& {p_v}\left( {u,v} \right) = \frac{1}{{{u^2} + {v^2} + 4}}\frac{\partial }{{\partial v}}\left[ {{u^2} + {v^2} + 4} \right] \cr
& {p_v}\left( {u,v} \right) = \frac{1}{{{u^2} + {v^2} + 4}}\left( {2v} \right) \cr
& {p_v}\left( {u,v} \right) = \frac{{2v}}{{{u^2} + {v^2} + 4}} \cr
& \cr
& {\text{Find the second partial derivatives }}{p_{uv}}\left( {u,v} \right){\text{ and }}{p_{vu}}\left( {u,v} \right){\text{ then}} \cr
& {p_{uv}}\left( {u,v} \right) = \frac{\partial }{{\partial v}}\left[ {\frac{{2u}}{{{u^2} + {v^2} + 4}}} \right] \cr
& {\text{use product rule}} \cr
& {p_{uv}}\left( {u,v} \right) = 2u\frac{\partial }{{\partial v}}\left[ {\frac{1}{{{u^2} + {v^2} + 4}}} \right] \cr
& {p_{uv}}\left( {u,v} \right) = 2u\left( { - \frac{1}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}}} \right)\frac{\partial }{{\partial v}}\left[ {{u^2} + {v^2} + 4} \right] \cr
& {p_{uv}}\left( {u,v} \right) = 2u\left( { - \frac{1}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}}} \right)\left( {2v} \right) \cr
& {p_{uv}}\left( {u,v} \right) = - \frac{{4uv}}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}} \cr
& and \cr
& {p_{vu}}\left( {u,v} \right) = \frac{\partial }{{\partial u}}\left[ {\frac{{2v}}{{{u^2} + {v^2} + 4}}} \right] \cr
& {\text{use product rule}} \cr
& {p_{vu}}\left( {u,v} \right) = 2v\frac{\partial }{{\partial u}}\left[ {\frac{1}{{{u^2} + {v^2} + 4}}} \right] \cr
& {p_{vu}}\left( {u,v} \right) = 2v\left( { - \frac{1}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}}} \right)\frac{\partial }{{\partial u}}\left[ {{u^2} + {v^2} + 4} \right] \cr
& {p_{vu}}\left( {u,v} \right) = 2v\left( { - \frac{1}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}}} \right)\left( {2u} \right) \cr
& {p_{vu}}\left( {u,v} \right) = - \frac{{4uv}}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}} \cr
& \cr
& {\text{Find the second partial derivatives }}{p_{uu}}\left( {u,v} \right){\text{ and }}{p_{vv}}\left( {u,v} \right){\text{ then}} \cr
& {p_{uu}}\left( {u,v} \right) = \frac{\partial }{{\partial u}}\left[ {\frac{{2u}}{{{u^2} + {v^2} + 4}}} \right] \cr
& {\text{by using the quotient rule}} \cr
& {p_{uu}}\left( {u,v} \right) = \frac{{\left( {{u^2} + {v^2} + 4} \right)\left( 2 \right) - 2u\left( {2u} \right)}}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}} \cr
& {p_{uu}}\left( {u,v} \right) = \frac{{2{u^2} + 2{v^2} + 8 - 4{u^2}}}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}} \cr
& {p_{uu}}\left( {u,v} \right) = \frac{{2{v^2} + 8 - 2{u^2}}}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}} \cr
& and \cr
& {\text{ }}{p_{vv}}\left( {u,v} \right) = \frac{\partial }{{\partial v}}\left[ {\frac{{2v}}{{{u^2} + {v^2} + 4}}} \right] \cr
& {\text{by using the quotient rule}} \cr
& {p_{vv}}\left( {u,v} \right) = \frac{{\left( {{u^2} + {v^2} + 4} \right)\left( 2 \right) - 2v\left( {2v} \right)}}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}} \cr
& {p_{vv}}\left( {u,v} \right) = \frac{{2{u^2} + 2{v^2} + 8 - 4{v^2}}}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}} \cr
& {p_{vv}}\left( {u,v} \right) = \frac{{2{u^2} + 8 - 2{v^2}}}{{{{\left( {{u^2} + {v^2} + 4} \right)}^2}}} \cr} $$