Answer
$-7$
Work Step by Step
Consider $\dfrac{\partial w}{\partial v}=\dfrac{\partial w}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial w}{\partial y}\dfrac{\partial y}{\partial v}$
and $(2x-\dfrac{y}{x^2})(-2)+(1/x)(1)=(2(u-2v+1)-(-2)\dfrac{(2u+v-2)}{(u-2v+1)^2})+(\dfrac{1}{(u-2v+1)})$
For the point $u=0,v=0$:
$\dfrac{\partial w}{\partial v}=(2(u-2v+1)-(-2)\dfrac{(2u+v-2)}{(u-2v+1)^2})+(\dfrac{1}{(u-2v+1)})=[2(0-0+1)-\dfrac{(0+0-2)}{(0-0+1)^2})](-2)+\dfrac{1}{1}=-7$
