Answer
$$\frac{\partial f(1,-1,-1)}{\partial x}=1$$
$$\frac{\partial f(1,-1,-1)}{\partial y}=1$$
$$\frac{\partial f(1,-1,-1)}{\partial z}=1$$
Work Step by Step
The partial derivative with respect to $x$ is:
$$\frac{\partial f(x,y,z)}{\partial x}=\frac{\partial }{\partial x}\left(\frac{x}{yz}\right)=\frac{1}{yz}\frac{\partial }{\partial x}(x)=\frac{1}{yz}\cdot1=\frac{1}{yz}$$
$y$ and $z$ are treated as a constant.
The partial derivative with respect to $y$ is:
$$\frac{\partial f(x,y,z)}{\partial y}=\frac{\partial }{\partial y}\left(\frac{x}{yz}\right)=\frac{x}{z}\frac{\partial }{\partial y}\left(\frac{1}{y}\right)=\frac{x}{z}\cdot\frac{-1}{y^2}=-\frac{x}{y^2z}$$
$x$ and $z$ are treated as a constant.
The partial derivative with respect to $z$ is:
$$\frac{\partial f(x,y,z)}{\partial z}=\frac{\partial }{\partial z}\left(\frac{x}{yz}\right)=\frac{x}{y}\frac{\partial }{\partial z}\left(\frac{1}{z}\right)=\frac{x}{y}\cdot\frac{-1}{z^2}=-\frac{x}{yz^2}$$
$x$ and $y$ are treated as a constant.
Now we will evaluate these partial derivatives in the given point $(x,y,z)=(1,-1,-1)$:
$$\frac{\partial f(1,-1,-1)}{\partial x}=\left.\frac{1}{yz}\right|_{(x,y,z)=(1,-1,-1)}=\frac{1}{-1\cdot(-1)}=1$$
$$\frac{\partial f(1,-1,-1)}{\partial y}=\left.-\frac{x}{y^2z}\right|_{(x,y,z)=(1,-1,-1)}=-\frac{1}{(-1)^2\cdot(-1)}=1$$
$$\frac{\partial f(1,-1,-1)}{\partial z}=\left.-\frac{x}{yz^2}\right|_{(x,y,z)=(1,-1,-1)}=-\frac{1}{-1\cdot(-1)^2}=1$$
