Answer
\begin{aligned} \frac{\partial x}{\partial u} &= -10 \sin( 10u-20v)\\
\frac{\partial g}{\partial v} &= 20\sin(10u-20v)
\end{aligned}
Work Step by Step
Given $$g(x, y)=\cos (x-y), \ \ \ \ x=3 u-5 v, \ \ \ y=-7 u+15 v$$ Since $$\frac{\partial g}{\partial x}=-\sin (x-y), \quad \frac{\partial g}{\partial y}=\sin (x-y)$$ and $$\begin{aligned} &\frac{\partial x}{\partial u}=3, \quad \frac{\partial x}{\partial v}=-5\\ &\frac{\partial y}{\partial u}=-7, \quad \frac{\partial y}{\partial v}=15 \end{aligned}$$ Then \begin{aligned} \frac{\partial g}{\partial u}&=\frac{\partial g}{\partial x} \frac{\partial x}{\partial u}+\frac{\partial g}{\partial y} \frac{\partial y}{\partial u}\\ &=-\sin (x-y) \frac{\partial x}{\partial u}+\sin (x-y) \frac{\partial y}{\partial u}\\
&= -\sin (x-y)(3)+\sin (x-y)(-7)\\ &=-10 \sin (x-y)\\
&= -10 \sin( 10u-20v)\\
\frac{\partial g}{\partial v}&=\frac{\partial g}{\partial x} \frac{\partial x}{\partial v}+\frac{\partial g}{\partial y} \frac{\partial y}{\partial v}\\ &=-\sin (x-y) \frac{\partial x}{\partial v}+\sin (x-y) \frac{\partial y}{\partial v}\\ &=-\sin (x-y)(-5)+\sin (x-y)(15)\\ &=20 \sin (x-y) \\
&= 20\sin(10u-20v)
\end{aligned}
