Answer
$${h_u}\left( {u,v} \right) = - \frac{{{v^2}}}{{2{{\left( {u - v} \right)}^2}}}\sqrt {\frac{{u-v}}{{uv}}} {\text{ and }}{h_v}\left( {u,v} \right) = \frac{{{u^2}}}{{2{{\left( {u - v} \right)}^2}}}\sqrt {\frac{{u-v}}{{uv}}} $$
Work Step by Step
$$\eqalign{
& h\left( {u,v} \right) = \sqrt {\frac{{uv}}{{u - v}}} \cr
& {\text{Calculate }}{h_u}\left( {u,v} \right),{\text{ treat }}v{\text{ as a constant}} \cr
& {h_u}\left( {u,v} \right) = \frac{\partial }{{\partial u}}\left[ {{{\left( {\frac{{uv}}{{u - v}}} \right)}^{1/2}}} \right] \cr
& {h_u}\left( {u,v} \right) = \frac{1}{2}{\left( {\frac{{uv}}{{u - v}}} \right)^{ - 1/2}}\frac{\partial }{{\partial u}}\left[ {\frac{{uv}}{{u - v}}} \right] \cr
& {h_u}\left( {u,v} \right) = \frac{1}{2}{\left( {\frac{{uv}}{{u - v}}} \right)^{ - 1/2}}\left( {\frac{{\left( {u - v} \right)v - uv\left( 1 \right)}}{{{{\left( {u - v} \right)}^2}}}} \right) \cr
& {h_u}\left( {u,v} \right) = \frac{1}{2}{\left( {\frac{{uv}}{{u - v}}} \right)^{ - 1/2}}\left( {\frac{{uv - {v^2} - uv}}{{{{\left( {u - v} \right)}^2}}}} \right) \cr
& {h_u}\left( {u,v} \right) = - \frac{{{v^2}}}{{2{{\left( {u - v} \right)}^2}}}\sqrt {\frac{{u-v}}{{uv}}} \cr
& \cr
& {\text{Calculate }}{h_v}\left( {u,v} \right),{\text{ treat }}u{\text{ as a constant}} \cr
& {h_v}\left( {u,v} \right) = \frac{\partial }{{\partial u}}\left[ {{{\left( {\frac{{u-v}}{{uv}}} \right)}^{1/2}}} \right] \cr
& {h_v}\left( {u,v} \right) = \frac{1}{2}{\left( {\frac{{uv}}{{u - v}}} \right)^{ - 1/2}}\frac{\partial }{{\partial u}}\left[ {\frac{{uv}}{{u - v}}} \right] \cr
& {h_v}\left( {u,v} \right) = \frac{1}{2}{\left( {\frac{{uv}}{{u - v}}} \right)^{ - 1/2}}\left( {\frac{{\left( {u - v} \right)u - uv\left( { - 1} \right)}}{{{{\left( {u - v} \right)}^2}}}} \right) \cr
& {h_v}\left( {u,v} \right) = \frac{1}{2}{\left( {\frac{{uv}}{{u - v}}} \right)^{ - 1/2}}\left( {\frac{{{u^2} - uv + uv}}{{{{\left( {u - v} \right)}^2}}}} \right) \cr
& {h_v}\left( {u,v} \right) = \frac{{{u^2}}}{{2{{\left( {u - v} \right)}^2}}}\sqrt {\frac{{u-v}}{{uv}}} \cr
& \cr
& {h_u}\left( {u,v} \right) = - \frac{{{v^2}}}{{2{{\left( {u - v} \right)}^2}}}\sqrt {\frac{{u-v}}{{uv}}} {\text{ and }}{h_v}\left( {u,v} \right) = \frac{{{u^2}}}{{2{{\left( {u - v} \right)}^2}}}\sqrt {\frac{{u-v}}{{uv}}} \cr} $$
