Let $$f(x, y, z) = \langle f_1(x,y,z), f_2(x,y,z), f_3(x,y,z)\rangle$$ be sufficiently differentiable and $\nabla \cdot f = 0$. Can anything be said about the expression $$(\nabla \times f) \cdot \nabla (f_i) - f \cdot \nabla \big((\nabla \times f)_i\big)$$ for $i = 1, 2, 3$?
I have tried computing this through brute force but it doesn't seem to simplify at all.
That equation is not true. Consider \begin{align} f=\begin{pmatrix}-y^2\\x^2\\0\end{pmatrix}\,. \end{align} This satisfies $\nabla\cdot f=0\,.$ Clearly, \begin{align} \nabla\times f&=\begin{pmatrix}0\\0\\2x+2y\end{pmatrix}\,,&\nabla f&=\begin{pmatrix}0&-2y&0\\2x&0&0\\0&0&0\end{pmatrix}\,,&\nabla(\nabla\times f)=\begin{pmatrix}0&0&0\\0&0&0\\2&2&0\end{pmatrix}\,. \end{align} The two terms in your equation are
\begin{align} (\nabla\times f)^\top(\nabla f)&=\begin{pmatrix}0\\0\\0 \end{pmatrix}\,,& f^\top(\nabla(\nabla\times f))&=\begin{pmatrix}0\\0\\\color{red}{2x^2} \end{pmatrix}\,. \end{align}