Answer
$$\frac{\partial}{\partial x}f(x,y)=3x^2yz^2;$$
$$\frac{\partial}{\partial y}f(x,y)=x^3z^2+2z;$$
$$\frac{\partial}{\partial z}f(x,y)=2x^3yz+2y.$$
Work Step by Step
Regard $y$ and $z$ as constant and differentiate with respect to $x$:
$$\frac{\partial}{\partial x}f(x,y)=\frac{\partial}{\partial x}(x^3yz^2+2yz)=yz^2\cdot3x^2+0=3x^2yz^2.$$
Regard $x$ and $z$ as constant and differentiate with respect to $y$:
$$\frac{\partial}{\partial y}f(x,y)=\frac{\partial}{\partial y}(x^3yz^2+2yz)=x^3z^2\cdot 1+2z\cdot 1=x^3z^2+2z.$$
Regard $x$ and $y$ as constant and differentiate with respect to $z$:
$$\frac{\partial}{\partial z}f(x,y)=\frac{\partial}{\partial z}(x^3yz^2+2yz)=x^3y\cdot 2z+2y\cdot1=2x^3yz+2y.$$
