I am trying to compute $f\nabla\times\mathbf{r}$ for $$ f(x,y,z) = \left(x^2+y^2+z^2\right)^{-\frac{3}{2}} \\ \mathbf{r} = x\mathbf{i}+y\mathbf{j}+z\mathbf{k} $$ I have computed $\nabla f\times\mathbf{r}$ which I determined to be the zero vector, but I do not yet have a solid grasp on the 'point free' notation of vector calculus. Here is what I have done so far:
Let $P =x^2+y^2+z^2$. $$ \begin{equation} f\nabla = P^{-\frac{3}{2}}\nabla = \left< P^{-\frac{3}{2}}\frac{\partial}{\partial x}, P^{-\frac{3}{2}}\frac{\partial}{\partial y}, P^{-\frac{3}{2}}\frac{\partial}{\partial z}\right> \end{equation} $$
I have interpreted the above as multiplying the scalar valued function by the operator $\nabla$ to give another operator $f\nabla$ that acts on scalar fields to give vector fields.
$$ \begin{align} f\nabla\times\mathbf{r} &= \begin{vmatrix} \mathbf{i} && \mathbf{j} && \mathbf{k} \\ P^{-\frac{3}{2}}\frac{\partial}{\partial x} && P^{-\frac{3}{2}}\frac{\partial}{\partial y} && P^{-\frac{3}{2}}\frac{\partial}{\partial z} \\ x && y && z\end{vmatrix} \\ &= \mathbf{i}\begin{vmatrix} P^{-\frac{3}{2}}\frac{\partial}{\partial y} && P^{-\frac{3}{2}}\frac{\partial}{\partial z} \\ y && z\end{vmatrix} - \cdots \end{align} $$
My question is: when evaluating each $2\times 2$ determinant above, should I interpret the expression $$ P^{-\frac{3}{2}}\frac{\partial}{\partial y}(z) $$ to mean the multiplication of $z$ and the operator $P^{-\frac{3}{2}}\frac{\partial}{\partial y}$ (giving another operator), or should I interpret it as the application of the operator $P^{-\frac{3}{2}}\frac{\partial}{\partial y}$ with $z$, giving a scalar? (in particular, $0$)