$r=(x,y,z)$ 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
 A: Following the determinant definition,
$$\operatorname{curl} r =  \left(\frac{\partial}{\partial y} z - \frac{\partial}{\partial z} y\right) \vec i - \left(\frac{\partial}{\partial x} z - \frac{\partial}{\partial z} x\right) \vec j +  \left(\frac{\partial}{\partial x} y - \frac{\partial}{\partial y} x\right)\vec k$$
Each of the six partial derivatives are zero, so the curl is $0 \vec i + 0 \vec j + 0 \vec k$, which is the zero vector.
A: Since $f(x,y,z)=\dfrac{x^2+y^2+z^2}2$ is such that ${\rm grad}\;f=(x,y,z)$, ${\rm curl}\;{\rm grad}\, f=0$
A: basically it uses that $\frac{\partial y}{\partial x}=0,\frac{\partial y}{\partial z}=0,\frac{\partial x}{\partial y}=0,\frac{\partial x}{\partial z}=0,\frac{\partial z}{\partial x}=0,\frac{\partial z}{\partial y}=0$ 
A: The reason the curl is $0$ is because $\mathbf{r}$ has continuous second-order partial derivatives. It's a known theorem. 
You should also note that this immediately implies $\mathbf{r}$ is a conservative field. 
A: 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.
