Find $\dfrac{\partial z}{\partial u}$ and $\dfrac{\partial z}{\partial v}$ if $z=f(x,y) =x^{2}+xy-y^{2}$ and $x=e^{2u+v}$ and $y=\ln \dfrac{ v}{u}.$

Solution The function $z=f(x,y)$ is a composite function of two independent variables $u$ and $v.$ So, we use Chain Rule II. $$\dfrac{\partial z}{\partial u}=\frac{\partial z}{\partial x}\frac{\partial x}{ \partial u}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial u}$$

Since $$\dfrac{\partial z}{\partial x}=2x+y\qquad \dfrac{ \partial z}{\partial y}=x-2y\qquad \dfrac{\partial x }{\partial u}=2e^{2u+v}\qquad \dfrac{\partial y}{\partial u} =\dfrac{\partial }{\partial u}( \ln v-\ln u) =-\dfrac{1}{u}$$

we have $$\begin{eqnarray*} \frac{\partial z}{\partial u}&=&\underset{\color{#0066A7}{\dfrac{\partial z}{\partial x}}}{ \underbrace{(2x+y)}}\underset{\color{#0066A7}{\dfrac{\partial x}{\partial u}}}{\underbrace{ (2e^{2u+v})}}+\underset{\color{#0066A7}{\dfrac{\partial z}{\partial y}}}{\underbrace{(x-2y)}}\underset{\color{#0066A7}{\dfrac{\partial y}{\partial u}}} {\underbrace{\left( -\dfrac{1}{u}\right) }}\\[4pt] &=&\left(2e^{2u+v}+\ln \dfrac{v}{u}\right) ( 2e^{2u+v}) -\left(e^{2u+v}-2\ln \dfrac{v}{u}\right) \left( \dfrac{1}{u}\right) \qquad \color{#0066A7}{{\hbox{\(x=e^{2u+v}; y=\ln \dfrac{v}{u}\)}}} \end{eqnarray*}$$

Similarly, since $\dfrac{\partial x}{\partial v}=e^{2u+v}$ and $\dfrac{ \partial y}{\partial v}=\dfrac{1}{v},$ we have $$\begin{eqnarray*} \dfrac{\partial z}{\partial v}&=&\underset{\color{#0066A7}{\dfrac{\partial z}{\partial x}}}{ \underbrace{(2x+y)}}\underset{\color{#0066A7}{\dfrac{\partial x}{\partial v}}}{\underbrace{ (e^{2u+v})}}+\underset{\color{#0066A7}{\dfrac{\partial z}{\partial y}}}{\underbrace{(x-2y)}} \underset{\color{#0066A7}{\dfrac{\partial y}{\partial v}}}{\underbrace{\left( \dfrac{1}{v} \right) }}\\[4pt] &=&\left( 2e^{2u+v}+\ln \dfrac{v}{u}\right) ( e^{2u+v}) +\left( e^{2u+v}-2~\ln \dfrac{v}{u}\right) \left( \dfrac{1}{v}\right) \qquad \color{#0066A7}{{\hbox{\(x=e^{2u+v}; y=\ln \dfrac{v}{u}\)}}}\qquad \end{eqnarray*}$$