I found in this material an identity that seems useful for me:
Suppose $\color{blue}{f}$ is a scalar function and $\color{blue}{u}$ is a vector function
$$\color{blue}{ \nabla \times (f\,u) = f\nabla \times u - u \times \nabla f}$$
Now, for make it clear, supposes: $$\color{blue}{\;f(x,y,z) = 2x+y}$$ $$\color{blue}{u(x,y,z) = x \hat x - 2y \hat y + z \hat z} $$ $$\color{blue}{ f(x,y,z)* u(z,y,z) = (2x^2 + xy) \hat x + (-2yx - 2y^2) \hat y + (2xz+yz) \hat z }$$
So $$\color{blue}{\nabla \times (f\,u) = curl (f(x,y,z)* u(z,y,z))}$$ $$\color{blue}{u \times \nabla f = u \times grad(u) = u \times [2,1,0]} $$
In the last expression above, $\color{blue}{\times}$ above is cross product.
However, $\color{blue}{f\nabla \times u\;}$ is a big mistery for me
I know a "cousin" (directional derivative) $$\color{blue}{f\nabla \cdot u = \partial f / \partial x \, u_x + \partial f / \partial y \, u_y + \partial f / \partial z \, u_z}$$
So I could imagine:
$$\color{blue}{f\nabla \times u = \begin{bmatrix}\hat x&\hat y & \hat z\\ \partial f / \partial x&\partial f / \partial y&\partial f / \partial z\\u_x&u_y&u_z \end{bmatrix}}$$
Is it real or am I crazy?
UPDATE
With the Arthur answer now I know the right interpretation:
$\color{blue}{f\nabla \times u\;}$ is $\color{blue}{f(\nabla \times u)\;}$
I've made
$$\color{blue}{\;f(x,y,z) = 2x}$$ $$\color{blue}{u(x,y,z) = -y \hat x + y \hat x + x \hat z} $$ $$\color{blue}{ f(x,y,z)* u(z,y,z) = (2xy) \hat x + (-2xy) \hat y + (2z^2) \hat z }$$
And it confers!
I also discover another reference here, page 30, identity D. The only difference is that appears $\;\color{blue}{+ \nabla f \times u \; }$ instead of $\;\color{blue}{- u \times \nabla f }$
The expression $f\nabla\times u$ means $f\cdot (\nabla\times u)$. The $f$ and the $\nabla$ do not interact directly here, and there isn't really any such thing as $f\nabla$.
(Well, technically, you can make it a thing that behaves just as the above. But I see little reason to do that, as it just increases the number of symbols whose meaning one has to remember, and it doesn't give us anything substantial in return.)