Answer
$$\frac{\partial f(1,-2,1)}{\partial x}=\frac{3}{\sqrt5}$$
$$\frac{\partial f(1,-2,1)}{\partial y}=-\frac{2}{\sqrt5}$$
$$\frac{\partial f(1,-2,1)}{\partial z}=-\frac{2}{\sqrt5}$$
Work Step by Step
The partial derivative with respect to $x$ is:
$$\frac{\partial f(x,y,z)}{\partial x}=\frac{\partial}{\partial x}(\sqrt{3x^2+y^2-2z^2})=\frac{1}{2\sqrt{3x^2+y^2-2z^2}}\frac{\partial}{\partial x}(3x^2+y^2-2z^2)=\frac{1}{2\sqrt{3x^2+y^2-2z^2}}\cdot6x=\frac{3x}{\sqrt{3x^2+y^2-2z^2}}$$
The partial derivative with respect to $y$ is:
$$\frac{\partial f(x,y,z)}{\partial y}=\frac{\partial}{\partial y}(\sqrt{3x^2+y^2-2z^2})=\frac{1}{2\sqrt{3x^2+y^2-2z^2}}\frac{\partial}{\partial y}(3x^2+y^2-2z^2)=\frac{1}{2\sqrt{3x^2+y^2-2z^2}}\cdot2y=\frac{y}{\sqrt{3x^2+y^2-2z^2}}$$
The partial derivative with respect to $z$ is:
$$\frac{\partial f(x,y,z)}{\partial z}=\frac{\partial}{\partial z}(\sqrt{3x^2+y^2-2z^2})=\frac{1}{2\sqrt{3x^2+y^2-2z^2}}\frac{\partial}{\partial z}(3x^2+y^2-2z^2)=\frac{1}{2\sqrt{3x^2+y^2-2z^2}}\cdot(-4z)=-\frac{2z}{\sqrt{3x^2+y^2-2z^2}}$$
Now we will evaluate these partial derivatives at the given point $(x,y,z)=(1,-2,1)$:
$$\frac{\partial f(1,-2,1)}{\partial x}=\left.\frac{3x}{\sqrt{3x^2+y^2-2z^2}}\right|_{(x,y,z)=(1,-2,1)}=\frac{3\cdot1}{\sqrt{3\cdot1^2+(-2)^2-2\cdot1^2}}=\frac{3}{\sqrt5}$$
$$\frac{\partial f(1,-2,1)}{\partial y}=\left.\frac{y}{\sqrt{3x^2+y^2-2z^2}}\right|_{(x,y,z)=(1,-2,1)}=\frac{-2}{\sqrt5}$$
$$\frac{\partial f(1,-2,1)}{\partial z}=-\left.\frac{2z}{\sqrt{3x^2+y^2-2z^2}}\right|_{(x,y,z)=(1,-2,1)}=-\frac{2}{\sqrt5}$$
