A formula for the gradient of the magnitude of a vector field $\mathbf{f}(x, y, z)$ is:
$$\nabla \|\mathbf{f}\| = \left(\frac{\mathbf{f}}{\|\mathbf{f}\|} \cdot \frac{\partial \mathbf{f}}{\partial x}, \frac{\mathbf{f}}{\|\mathbf{f}\|} \cdot \frac{\partial \mathbf{f}}{\partial y}, \frac{\mathbf{f}}{\|\mathbf{f}\|} \cdot \frac{\partial \mathbf{f}}{\partial z}\right)$$
Can this be written in a form that doesn't mention coordinates explicitly? (i.e., using nablas, dot products, etc.)
$$ \mathrm{d} \|f\|^2 = 2 \|f\| \, \mathrm{d} \|f\|$$ $$ \mathrm{d} \|f\|^2 = \mathrm{d}(f \cdot f) = f \cdot \mathrm{d}f + \mathrm{d}f \cdot f$$
The meaning of most objects involved is clear; e.g.
and so forth. The tricky part is trying to figure out what kind of object $\mathrm{d}f$ is, and what dot products with it mean: in coordinates, $\mathrm{d}f$ is a matrix whose rows are the derivatives of the components of $f$, and the dot product of a matrix and a column vector would be the row vector whose entries are the dot products of then individual columns of the matrix with the vector. And so
$$ \mathrm{d}\|f\| = \frac{f}{\|f\|} \cdot \mathrm{d}f = \frac{f^T}{\|f\|} \mathrm{d}f$$
where $f^T$ means the covector we get by transposing $f$ with respect to the dot product, and the product is the ordinary product of a covector with a linear operator. (i.e. in coordinates, the usual product of a row vector with a matrix)
It's been a long time since I've used nabla-notation for this, but I think they still use $\nabla f$ for $\mathrm{d}f$.