Answer
$$\frac{dw}{dt}=e^t(\sin(\pi-t)-\cos(\pi-t))$$
When $t=0$
$$\frac{dw}{dt}=1$$
Work Step by Step
Using the Chain rule we have:
$$\frac{dw}{dt}=\frac{\partial w}{\partial x}\frac{dx}{dt}+\frac{\partial w}{\partial y}\frac{dy}{dt}=\frac{\partial}{\partial x}(x\sin y)\frac{d}{dt}(e^t)+\frac{\partial}{\partial y}(x\sin y)\frac{d}{dt}(\pi-t)=\sin y\cdot e^t+x\cos y\cdot(-1)=e^t\sin y-x\cos y$$
Expressing this in terms of $t$ we have:
$$\frac{dw}{dt}=e^t\sin(\pi-t)-e^t\cos(\pi-t)=e^t(\sin(\pi-t)-\cos(\pi-t))$$
When $t=0$ we have:
$$\frac{dw}{dt}=\left.e^t(\sin(\pi-t)-\cos(\pi-t))\right|_{t=0}=e^0(\sin(\pi-0)-\cos(\pi-0))=1\cdot(0-(-1))=1$$
