Find $\nabla \cdot (f\textbf r)$ and $\nabla \times (f\textbf r)$ of the function $f(x,y,z) = (x^2+y^2)\log(1-z)$ I have been given the function $f(x,y,z) = (x^2+y^2)\log(1-z)$ and I need to find the divergence $\nabla \cdot (f\textbf r)$ and curl $\nabla \times (f\textbf r)$ where $\textbf r$ is the position vector.
I understand that $$\nabla \cdot F = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}$$
and that $$\nabla \times F = \begin{bmatrix}
                                i & j & k \\
                                \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
                                F_x & F_y & F_z \\
                            \end{bmatrix}
                            $$
I just don't know how to proceed with this question.

You multiply the function by the position vector and then just calculate the derivatives...
[\begin{array}{l}
\vec r = \left( {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right)\\
f(x,y,z) = \left( {{x^2} + {y^2}} \right)\log (1 - z) = {x^2}\log (1 - z) + {y^2}\log (1 - z)\\
f(x,y,z)\vec r = \vec F = \left( {\begin{array}{*{20}{c}}
{{x^3}\log (1 - z) + x{y^2}\log (1 - z)}\\
{{x^2}y\log (1 - z) + {y^3}\log (1 - z)}\\
{{x^2}z\log (1 - z) + {y^2}z\log (1 - z)}
\end{array}} \right)\\
\nabla  \bullet \vec F = {\partial _x}{F_x} + {\partial _y}{F_y} + {\partial _z}{F_z} = ...\\
\nabla  \times \vec F = \left| {\begin{array}{*{20}{c}}
{\hat i}&{\hat j}&{\hat k}\\
{{\partial _x}}&{{\partial _y}}&{{\partial _z}}\\
{{F_x}}&{{F_y}}&{{F_z}}
\end{array}} \right| = \left( {\begin{array}{*{20}{c}}
{{\partial _y}{F_z} - {\partial _z}{F_y}}\\
{{\partial _z}{F_x} - {\partial _x}{F_z}}\\
{{\partial _x}{F_y} - {\partial _y}{F_x}}
\end{array}} \right) = ...
\end{array}]
