For any scalar function
\[f(x,y,x)\]
with continuous partial derivatives \[\mathbf{\nabla} \times (f \nabla f) =0\]
Proof
\[\begin{equation} \begin{aligned} \mathbf{\nabla} \times (f \nabla f) &=
(\frac{\partial }{\partial x} \mathbf{i} + \frac{\partial }{\partial y} \mathbf{j} + \frac{\partial }{\partial z} \mathbf{k}) \times (f \frac{\partial f}{\partial x} \mathbf{i} + f \frac{\partial f}{\partial y} \mathbf{j} + f \frac{\partial f}{\partial z} \mathbf{k}) \\ &=
(\frac{\partial}{\partial y}(f \frac{\partial f}{\partial z}) - \frac{\partial}{\partial z}(f \frac{\partial f}{\partial y})\mathbf{i} +
(\frac{\partial}{\partial z}(f \frac{\partial f}{\partial x}) - \frac{\partial}{\partial x}(f \frac{\partial f}{\partial z})\mathbf{j} +
(\frac{\partial}{\partial x}(f \frac{\partial f}{\partial y}) - \frac{\partial}{\partial y}(f \frac{\partial f}{\partial x})\mathbf{k} \\ &=
(\frac{\partial f}{\partial y} \frac{\partial f}{\partial z} +f \frac{\partial^2 f}{\partial y \partial z} - \frac{\partial f}{\partial z} \frac{\partial f}{\partial y} - f \frac{\partial^2 f}{\partial z\partial y})\mathbf{i} +
(\frac{\partial f}{\partial z} \frac{\partial }{\partial x}+ f \frac{\partial^2 f}{\partial z \partial x} - \frac{\partial f}{\partial x}\frac{\partial f}{\partial z} - f \frac{\partial^2 f}{\partial x \partial z}) \mathbf{j} \\ & +
(\frac{\partial f}{\partial x} \frac{\partial f}{\partial y} +f \frac{\partial^2 f}{\partial x \partial y}- \frac{\partial f}{\partial y} \frac{\partial^2 f}{\partial y \partial x})\mathbf{k} \\ &= 0 \end{aligned} \end{equation}
\]