showing / proving curl identity $\nabla \times \left( \frac{1}{r^2} \hat r \right) = 0$ OK, I have to show the following: 
$$ \nabla \times \left( \frac{1}{r^2}  \hat r \right) = 0$$
This should be pretty easy, but I wanted to be sure I was doing this correctly. 
I set up the matrix: 
$$
        \begin{bmatrix}
        \hat r & \hat \theta & \hat \phi \\
        \frac{\partial}{\partial r}  & \frac{\partial}{\partial \theta} & \frac{\partial}{\partial \phi} \\
        \frac{1}{r^2} & 0 & 0 \\
        \end{bmatrix}
=\left(\frac{\partial}{\partial \theta}(0)-\frac{\partial}{\partial \phi}(0)\right)\hat r-\left(\frac{\partial}{\partial r}(0)-\frac{\partial}{\partial \phi}(\frac{1}{r^2})\right)\hat \theta-\left(\frac{\partial}{\partial r}(0)-\frac{\partial}{\partial \theta}(\frac{1}{r^2})\right)\hat \phi$$
which leaves me with 0 because $\frac{\partial}{\partial \theta}(\frac{1}{r^2})$ and $\frac{\partial}{\partial \phi}(\frac{1}{r^2})$ are both zero. 
This is correct, yes? I know this is ridiculously simple a problem but I want to make sure I did not forget everything I learned last semester. (Also, I was curious if there is a more rigorous proof, tho this is for a phys and not a math class). 
Edit: BTW this is in spherical (I think -- the assignment uses $\hat r$ so I am going with that). 
 A: Actually we can check this is true using the following two facts:


*

*curl of a gradient field is zero.

*cross product of two parallel vector fields is zero.
I am assuming your $\hat{r} = (x,y,z)$, and $r= \sqrt{x^2+y^2+z^2}$, then it is not hard to check that 
$$
\frac{1}{r^2} \hat{r} = \frac{\nabla r}{r}.
$$
Now using the product rule for curl, a scalar function $f$ and a vector field $\hat{g}$:
$$
\nabla \times (f\hat{g}) = \nabla f \times \hat{g} + f\nabla\times \hat{g}.
$$
We have
$$
\nabla \times  \left(\frac{1}{r^2} \hat{r}\right)= \nabla \times \left(\frac{\nabla r}{r}\right) = \nabla \times (\nabla r) \frac{1}{r} + 
\nabla \left(\frac{1}{r}\right)\times \nabla r = 0 - \frac{1}{r^2}\nabla r\times \nabla r=0.
$$
A: Any vector field that can be expressed in the form $f(r)\mathbf{\hat r}$ must necessarily have zero curl (where the function is smooth, at least).
This can be seen by noting that, if you have a scalar field $g(r)$, and you take its gradient, you get $g'(r)\mathbf{\hat r}$. As such, with $f(r)=g'(r)$, you get the vector field. Now, you have $$\nabla \times f(r)\mathbf{\hat r} = \nabla \times \nabla g(r) = 0$$
In particular, for $f(r)=\frac1{r^2}$, you have $g(r)=-\frac1r$.
Note that this only applies if the function is independent of $\theta$ and $\phi$.
A: Set $U(r) = \frac{1}{r^2}$; then this problem is a special case of the following, which also appears here.
If what you really want to do is show that $\nabla \times U(r) \hat r = 0$ for spherically stmmetric functions $U(r)$, then perhaps the easiest thing to do is use the identity
$\nabla \times (f \hat V) = \nabla f \times \hat V + f \nabla \times \hat V, \tag{1}$
a standard result from vector calculus, which can easily be derived from the well-known "determinant formula" for $\nabla \times$:
$\nabla \times \hat Y = \begin{bmatrix} \hat {\mathbf i} & \hat {\mathbf j} & \hat {\mathbf k} \\ \frac{\partial}{\partial x} &  \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \hat Y_x & \hat Y_y & \hat Y_z \end{bmatrix}. \tag{2}$
If we set $\hat Y = f \hat V$ in (2) and grind, we get
$\nabla \times f \hat V = (\frac{\partial}{\partial y} (f \hat V_z) - (\frac{\partial}{\partial z} (f \hat V_y))\hat {\mathbf i} + (\frac{\partial}{\partial z} (f \hat V_x) - (\frac{\partial}{\partial x} (f \hat V_z))\hat {\mathbf j}$
$+ (\frac{\partial}{\partial x} (f \hat V_y) - (\frac{\partial}{\partial y} (f \hat V_x))\hat {\mathbf k}, \tag{3}$
or
$\nabla \times f \hat V = ((f_y \hat V_z + f \hat V_{z, y}) -  (f_z \hat V_y + f \hat V_{y, z}))\hat {\mathbf i} + ((f_z \hat V_x + f \hat V_{x, z}) - (f_x \hat V_z + f \hat V_{z, x}))\hat {\mathbf j}$
$+ ((f_x \hat V_y + f \hat V_{y, x}) - (f_y \hat V_x + f \hat V_{x, y}))\hat {\mathbf k}, \tag{4}$
after which some simple algebraic re-arrangement yields (1).  For example, the coefficient of $\hat{\mathbf i}$ is
$(f_y \hat V_z - f_z \hat V_y) + f(\hat V_{z, y} - \hat V_{y, z}). \tag{5}$
In (3), (4), and (5), I have used subscript notation for partials, e.g. $f_x = \frac{\partial f}{\partial x}$ and $\hat V_{x,y} = \frac{\partial V_x}{\partial y}$ etc.  This formula may also be found in this wikipedia article.
Applying (1) to $U(r) \hat r$, we have
$\nabla \times U(r) \hat r = \nabla U(r) \times \hat r + U(r) \nabla \times \hat r. \tag{6}$
Now
$\nabla \times \hat r = 0, \tag{7}$
which is easy to see by direct calculation, using (2) if you like, so we are left with
$\nabla \times U(r) \hat r = \nabla U(r) \times \hat r; \tag{8}$
but I claim that for any function $U(r)$ which only depends on $r$, $\nabla U(r)$ is in fact collinear with $\hat r$; to see this, write
$U(r) = U((x^2 + y^2 + z^2)^{\frac{1}{2}}), \tag{9}$
so that for example
$\frac{\partial U}{\partial x}(r) = \frac{dU}{dr}(r) \frac{x}{r}, \tag{10}$
by the chain rule, since $\frac{\partial r}{\partial x} = \frac{x}{r}$.  Similar expressions hold for the $y$ and $z$ derivatives as well.  Now (10), when combined with its $y$ and $z$ counterparts, gives
$\nabla U(r) = r^{-1}\frac{dU}{dr}(r) \hat r, \tag{11}$,
and substituting this in (8) yields
$\nabla \times U(r) \hat r = r^{-1}\frac{dU}{dr}(r) \hat r \times \hat r = 0, \tag{12}$
since $\hat r \times \hat r$ vanishes identically.  Thus we see that
$\nabla \times U(r) \hat r = 0, \tag{13}$
as was required.  QED.
CAVEAT:  One should of course take care applying these or any derivative formula to a function or vector field at the origin $(0, 0, 0)$ where $r = 0$, since $\sqrt{x^2 + y^2 + z^2}$ is continuous but not differentiable there. But if $U(r)$ is sufficiently smooth everywhere, then I do b'lieve she'll fly, Wilbur!
Hope this helps.  Cheers!
and as always
Fiat Lux!
