The textbook example gives an implicitly defined function of $$x^3+y^3+z^3+6xyz=1$$ It then walks through the first step of partial differentiation of z with respect to $x$:
$$3x^2+3z^2\frac{\partial z}{\partial x}+6yz+6xy\frac{\partial z}{\partial x}=0$$
Why does the 6xyz part become two things? If I'm partial differentiating with respect to x, why isn't it just $6yz\frac{\partial z}{\partial x}$ ?
So, it seems that the given relation allows us to define $z = z(x,y)$...So differentiating with respect to $x$ gives in fact $$ 3x^2 + 3 \frac{\partial z}{\partial x} z^2+6yz +6xy \frac{\partial z}{\partial x} = 0 $$
The term you are asking for is just the derivative of a product: $$ (6xyz)' = 6y(x'_x z + x z'_x) = 6y(z+x z'_x) $$