Answer
$$\frac{\partial f(\pi,0)}{\partial x}=-1$$
$$\frac{\partial f(\pi,0)}{\partial y}=0$$
Work Step by Step
The partial derivative with the respect to $x$ is:
$$\frac{\partial f(x,y)}{\partial x}=\frac{\partial }{\partial x}(e^y\sin x)=e^y\frac{\partial }{\partial x}(\sin x)=e^y\cos x,$$
because $y$ is treated as a constant.
The partial derivative with the respect to $y$ is:
$$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y}(e^y\sin x)=\sin x\frac{\partial }{\partial y }(e^y)=e^y\sin x,$$
because now $x$ is treated as a constant.
Now we will evaluate these partial derivatives in the given point $(x,y)=(\pi,0)$:
$$\frac{\partial f(\pi,0)}{\partial x}=\left.e^y\cos x\right|_{(x,y)=(\pi,0)}=e^0\cos\pi=1\cdot(-1)=-1$$
$$\frac{\partial f(\pi,0)}{\partial y}=\left.e^y\sin x\right|_{(x,y)=(\pi,0)}=e^0\sin \pi=1\cdot0=0$$
