Answer
$$\frac{\partial f(x,y,z)}{\partial x}=6xy-5yz$$
$$\frac{\partial f(x,y,z)}{\partial y}=3x^2-5xz+10z^2$$
$$\frac{\partial f(x,y,z)}{\partial z}=20yz-5xy$$
Work Step by Step
The partial derivative with respect to $x$ is:
$$\frac{\partial f(x,y,z)}{\partial x}=\frac{\partial }{\partial x}(3x^2y-5xyz+10yz^2)=3y\frac{\partial }{\partial x}(x^2)-5yz\frac{\partial f}{\partial x}(x)+\frac{\partial }{\partial x}(10yz^2)=3y\cdot2x-5yz\cdot1+0=6xy-5yz$$
Here are $y$ and $z$ treated as a constant.
The partial derivative with respect to $y$ is:
$$\frac{\partial f(x,y,z)}{\partial y}=\frac{\partial }{\partial y}(3x^2y-5xyz+10yz^2)=3x^2\frac{\partial }{\partial y}(y)-5xz\frac{\partial }{\partial y}(y)+10z^2\frac{\partial }{\partial y}(y)=3x^2\cdot1-5xz\cdot1+10z^2\cdot1=3x^2-5xz+10z^2$$
Here are $x$ and $z$ treated as a constant.
The partial derivative with respect to $z$ is:
$$\frac{\partial f(x,y,z)}{\partial z}=\frac{\partial }{\partial z}(3x^2y-5xyz+10yz^2)=\frac{\partial }{\partial z}(3x^2y)-5xy\frac{\partial }{\partial z}(z)+10y\frac{\partial }{\partial z}(z^2)=0-5xy\cdot1+10y\cdot2z=20yz-5xy$$
Here are $x$ and $y$ treated as a constant.