Answer
$$\frac{\partial f(0,\frac{\pi}{2},-4)}{\partial x}=0$$
$$\frac{\partial f(0,\frac{\pi}{2},-4)}{\partial y}=0$$
$$\frac{\partial f(0,\frac{\pi}{2},-4)}{\partial z}=1$$
Work Step by Step
The partial derivative with respect to $x$ is:
$$\frac{\partial f(x,y,z)}{\partial x}=\frac{\partial}{\partial x}(z\sin(x+y))=z\cos(x+y)\frac{\partial}{\partial x}(x+y)=z\cos(x+y)\cdot1=z\cos(x+y)$$
The partial derivative with respect to $y$ is:
$$\frac{\partial f(x,y,z)}{\partial y}=\frac{\partial}{\partial y}(z\sin(x+y))=z\cos(x+y)\frac{\partial}{\partial y}(x+y)=z\cos(x+y)\cdot1=z\cos(x+y)$$
The partial derivative with respect to $z$ is:
$$\frac{\partial f(x,y,z)}{\partial z}=\frac{\partial}{\partial z}(z\sin(x+y))=\sin(x+y)$$
Now we will evaluate these partial derivatives at the given point $(x,y,z)=(0,\frac{\pi}{2},-4)$:
$$\frac{\partial f(0,\frac{\pi}{2},-4)}{\partial x}=\left.z\cos(x+y)\right|_{(x,y,z)=(0,\pi/2,-4)}=-4\cos(0+\frac{\pi}{2})=-4\cdot0=0$$
$$\frac{\partial f(0,\frac{\pi}{2},-4)}{\partial y}=\left.z\cos(x+y)\right|_{(x,y,z)=(0,\pi/2,-4)}=-4\cos(0+\frac{\pi}{2})=-4\cdot0=0$$
$$\frac{\partial f(0,\frac{\pi}{2},-4)}{\partial z}=\left.\sin(x+y)\right|_{(x,y,z)=(0,\pi/2,-4)}=\sin(0+\frac{\pi}{2})=1$$