Answer
$\frac{{\partial f}}{{\partial u}} = - 2v{{\rm{e}}^{u - v}}$
$\frac{{\partial f}}{{\partial v}} = 2\left( {v - 1} \right){{\rm{e}}^{u - v}}$
Work Step by Step
We have $f\left( {x,y} \right) = \left( {x - y} \right){{\rm{e}}^x}$, where $x=u-v$ and $y=u+v$.
The partial derivatives are
$\frac{{\partial f}}{{\partial x}} = {{\rm{e}}^x} + \left( {x - y} \right){{\rm{e}}^x} = \left( {1 + x - y} \right){{\rm{e}}^x}$, ${\ \ \ }$ $\frac{{\partial f}}{{\partial y}} = - {{\rm{e}}^x}$
$\frac{{\partial x}}{{\partial u}} = 1$, ${\ \ \ }$ $\frac{{\partial y}}{{\partial u}} = 1$
$\frac{{\partial x}}{{\partial v}} = - 1$, ${\ \ \ }$ $\frac{{\partial y}}{{\partial v}} = 1$
Using the Chain Rule, Eq. (2) and Eq. (3) of Section 15.6, we have
$\frac{{\partial f}}{{\partial u}} = \frac{{\partial f}}{{\partial x}}\frac{{\partial x}}{{\partial u}} + \frac{{\partial f}}{{\partial y}}\frac{{\partial y}}{{\partial u}}$
$\frac{{\partial f}}{{\partial v}} = \frac{{\partial f}}{{\partial x}}\frac{{\partial x}}{{\partial v}} + \frac{{\partial f}}{{\partial y}}\frac{{\partial y}}{{\partial v}}$
So,
(1) ${\ \ \ }$ $\frac{{\partial f}}{{\partial u}} = \left( {1 + x - y} \right){{\rm{e}}^x} - {{\rm{e}}^x} = \left( {x - y} \right){{\rm{e}}^x}$
(2) ${\ \ \ }$ $\frac{{\partial f}}{{\partial v}} = - \left( {1 + x - y} \right){{\rm{e}}^x} - {{\rm{e}}^x} = - \left( {2 + x - y} \right){{\rm{e}}^x}$
Since $x=u-v$ and $y=u+v$, we get $x-y=-2v$
Substituting $x=u-v$ and $x-y=-2v$ in equation (1) and (2) we obtain $\frac{{\partial f}}{{\partial u}}$ and $\frac{{\partial f}}{{\partial v}}$ in terms of $u$ and $v$:
$\frac{{\partial f}}{{\partial u}} = - 2v{{\rm{e}}^{u - v}}$
$\frac{{\partial f}}{{\partial v}} = - \left( {2 - 2v} \right){{\rm{e}}^{u - v}} = 2\left( {v - 1} \right){{\rm{e}}^{u - v}}$
