Answer
$$\frac{\partial f(-2,1,2)}{\partial x}=0$$
$$\frac{\partial f(-2,1,2)}{\partial y}=-2$$
$$\frac{\partial f(-2,1,2)}{\partial z}=-7$$
Work Step by Step
The partial derivatives with respect to $x$ is:
$$\frac{\partial f(x,y,z)}{\partial x}=\frac{\partial}{\partial x}(x^2y^3+2xyz-3yz)=y^3\frac{\partial }{\partial x}(x^2)+2yz\frac{\partial }{\partial x}(x)-\frac{\partial }{\partial x}(3yz)=y^3\cdot2x+2yz\cdot1+0=2xy^3+2yz$$
$y$ and $z$ are treated as a constant.
The partial derivative with respect to $y$ is:
$$\frac{\partial f(x,y,z)}{\partial y}=\frac{\partial }{\partial y}(x^2y^3+2xyz-3yz)=x^2\frac{\partial }{\partial y}(y^3)+2xz\frac{\partial }{\partial y}(y)-3z\frac{\partial }{\partial y}(y)=x^2\cdot3y^2+2xz\cdot1-3z\cdot1=3x^2y^2+2xz-3z$$
$x$ and $z$ are treated as a constant.
The partial derivatives with respect to $z$ is:
$$\frac{\partial f(x,y,z)}{\partial z}=\frac{\partial }{\partial z}(x^2y^3+2xyz-3yz)=\frac{\partial }{\partial z}(x^2y^3)+2xy\frac{\partial}{\partial z}(z)-3y\frac{\partial }{\partial z}(z)=0+2xy\cdot1-3y\cdot1=2xy-3y$$
$x$ and $y$ are treated as a constant.
Now we will evaluate these partial derivatives in the given point $(x,y,z)=(-2,1,2)$:
$$\frac{\partial f(-2,1,2)}{\partial x}=\left.(2xy^3+2yz)\right|_{(x,y,z)=(-2,1,2)}=2\cdot(-2)\cdot1^3+2\cdot1\cdot2=-4+4=0$$
$$\frac{\partial f(-2,1,2)}{\partial y}=\left.(3x^2y^2+2xz-3z)\right|_{(x,y,z)=(-2,1,2)}=3\cdot(-2)^2\cdot1^2+2\cdot(-2)\cdot2-3\cdot2=12-8-6=-2$$
$$\frac{\partial f(-2,1,2)}{\partial z}=\left.(2xy-3y)\right|_{(x,y,z)=(-2,1,2)}=2\cdot(-2)\cdot1-3\cdot1=-7$$
