Answer
$$\frac{\partial f(2,\frac{\pi}{4})}{\partial x}=0$$
$$\frac{\partial f(2,\frac{\pi}{4})}{\partial y}=0$$
Work Step by Step
To solve both partial derivatives we will use chain rule.
The partial derivative with the respect to $x$ is:
$$\frac{\partial f(x,y)}{\partial x}=\frac{\partial }{\partial x}(\sin xy)=\cos xy\frac{\partial }{\partial x}(xy)=y\cos xy$$
because $y$ is treated as a constant.
The partial derivative with the respect to $y$ is:
$$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y}(\sin xy)=\cos xy\frac{\partial}{\partial y}(xy)=x\cos xy$$
because now $x$ is treated as a constant.
Now we will evaluate these partial derivatives in the given point $(x,y)=(2,\frac{\pi}{4})$:
$$\frac{\partial f(2,\frac{\pi}{4})}{\partial x}=\left.y\cos xy\right|_{(x,y)=(2,\frac{\pi}{4})}=\frac{\pi}{4}\cos2\cdot\frac{\pi}{4}=\frac{\pi}{4}\cdot\cos\frac{\pi}{2}=\frac{\pi}{4}\cdot0=0$$
$$\frac{\partial f(2,\frac{\pi}{4})}{\partial y}=\left.x\cos xy\right|_{(x,y)=(2,\frac{\pi}{4})}=2\cos2\frac{\pi}{4}=2\cos\frac{\pi}{2}=2\cdot0=0$$
