$r=(x,y,z)$ prove that $\mathrm{curl}\; r = 0$ 
Example $\bf 84\,\,\,$ Let $\,\mathbf r=(x,y,z)$ and $r=|\!\,\mathbf r|=\sqrt{x^2+y^2+z^2}$. Then $$\operatorname{div}\mathbf r=
\dfrac{\partial x}{\partial x}
+
\dfrac{\partial y}{\partial y}
+
\dfrac{\partial z}{\partial z}
=3; \\
\operatorname{curl}\mathbf r=
\left|\begin{matrix}
\mathbf i & \mathbf j & \mathbf k \\
\tfrac{\partial}{\partial x} & \tfrac{\partial}{\partial y} & \tfrac\partial{\partial z}\\
x & y & z 
\end{matrix}\right|=\rlap{\rlap{0}\rlap00}0.$$

I fail to see how this equals zero
Is $\tfrac{dz}{dy} - \tfrac{dy}{dz} = 0$ and same for other terms too?
Thanks

One can also notice that this is a spherically symmetric problem, so it is sufficient to evaluate only on one axis, e.g. on $x$ axis. What follows also applies for a more general case where $\mathbf r = (x,y,z)f(\sqrt{x^2+y^2+z^2})$.
Along $x$-axis, due to the symmetry, for each component only the derivative along $x$-axis is non-vanishing. Moreover, since only $x$-component is nonzero along the $x$-axis, only $x$-derivative of $x$-component remains. However, there is no such derivative in the curl operator, therefore curl is zero along the $x$-axis.
Because of the spherical symmetry, the curl is also zero anywhere else.
