Answer
$$\dfrac{\partial w}{\partial r}=2; \\\dfrac{\partial w}{\partial s}=2-\pi$$
Work Step by Step
The partial derivatives at $r=\pi, s=0$ are as follows:
$$x=r+sin(s)=\pi ;\\ \dfrac{dx}{dr}=1, \dfrac{dx}{ds}=\ cos s=1; \\ y=rs=0$$
Now, $$y_r=r \implies y_r=0, y_s=r=\pi$$
$$w=\sin(2x-y)=0; \\ w_x=2cos(2x-y)=2, w_y=-\cos(2x-y)=-1$$
Now, $$w_r=w_x \dfrac{dx}{dr}+w_y \dfrac{dy}{dr}=2(1)-1(0)=2$$ and $$\dfrac{\partial w}{\partial s}=w_x \dfrac{dx}{ds}+w_y \times \dfrac{dy}{ds}=2(1)-1(\pi)=2-\pi$$
