As John Hughes already mentioned, we require $\nabla \cdot \vec J=0$. Under that restriction, we proceed.
Since the curl of the gradient is zero ($\nabla \times \nabla \Phi=0$), then if
$$\nabla \times \vec B =\mu_0 \vec J$$
for the magnetic field $\vec B$, then we also have
$$\nabla \times (\vec B+\nabla \Phi) =\mu_0 \vec J$$
for any (smooth) scalar field $\Phi$. This means that there is not a unique solution to the problem since $\vec B +\nabla \Phi$ is also a solution for any (smooth) $\Phi$.
However, if we also specify the divergence of the magnetic field (we know that it is zero), then we can pursue a unique solution. For example,
$$\nabla \times \nabla \times \vec B =-\mu_0 \nabla \times \vec J$$
whereupon using the vector identity $\nabla \times \nabla \times \vec B= \nabla ( \nabla \cdot \vec B)-\nabla^2 \vec B$ and exploiting $\nabla \cdot \vec B=0$ gives
$$\nabla^2 \vec B=-\mu_0 \nabla \times \vec J$$
which has solution
$$\vec B(\vec r)=\mu_0 \int_V \frac{\nabla' \times \vec J(\vec r')}{4\pi |\vec r-\vec r'|}dV'$$
where the volume integral extends over all space where $\vec J=\ne 0$. We can integrate by parts in three dimensions by using the vector product rule identity $\nabla \times (\Phi \vec A) = \Phi \nabla \times \vec A+\nabla \Phi \times \vec A$ to write
$$\begin{align}
\vec B(\vec r)&=\mu_0 \int_V \frac{\nabla' \times \vec J(\vec r')}{4\pi |\vec r-\vec r'|}dV'\\\\
&=\mu_0 \int_V \left(\nabla' \times \left(\frac{\vec J(\vec r')}{4\pi |\vec r-\vec r'|}\right) -\nabla' \left(\frac{1}{4\pi |\vec r-\vec r'|}\right) \times \vec J(\vec r') \right)dV'\\\\
&=\mu_0 \oint_S \frac{\hat n' \times \vec J(\vec r')}{4\pi |\vec r-\vec r'|}dS'-
\mu_0 \int_V \nabla' \left(\frac{1}{4\pi |\vec r-\vec r'|}\right) \times \vec J(\vec r') dV'
\end{align}$$
Now, we may extend the integration region to all of space. Then, if $\vec J=0$ outside a finite region, then the surface integral vanishes and we have
$$\begin{align}
\vec B(\vec r)&= -\mu_0
\int_V \nabla' \left(\frac{1}{4\pi |\vec r-\vec r'|}\right) \times \vec J(\vec r') dV'\\\\
&=\mu_0 \int_V \nabla \left(\frac{1}{4\pi |\vec r-\vec r'|}\right) \times \vec J(\vec r') dV'\\\\
&=\nabla \times \left( \mu_0 \int_V \frac{\vec J(\vec r')}{4\pi |\vec r-\vec r'|}dV' \right)
\end{align}$$
where the first equality is effectively the Biot-Savart law and the last equality reveals that $\vec B =\nabla \times \vec A$ for the vector potential
$$\vec A(\vec r) = \mu_0 \int_V \frac{\vec J(\vec r')}{4\pi |\vec r-\vec r'|}dV'$$
Yes, there's a more elegant way! It uses the language of differential forms, which has replaced the 19th-century language of gradients, divergences, and curls in modern geometry. You can appreciate the simplicity of this language even before learning how to read it:
For any 1-form $A$,
$$\begin{align*}
(\star d)(\star d)A & = (\star d \star)dA \\
\operatorname{curl} \operatorname{curl} A & = d^\dagger dA
\end{align*}$$
Recalling that $\Delta = d d^\dagger + d^\dagger d$, we see that
$$\begin{align*}
\operatorname{curl} \operatorname{curl} A & = -d d^\dagger A + \Delta A \\
& = d(\star d \star)A + \Delta A \\
& = \operatorname{grad} \operatorname{div} A + \Delta A
\end{align*}$$
This is the identity you wanted to prove, where $-\Delta$ is the vector Laplacian.
My favorite place to learn about differential forms is in Chapters 4 and 5 of Gauge Fields, Knots, and Gravity by John Baez and Javier Muniain.
Here's a rough glossary that should help you move between the language of differential forms and the old language of vector calculus. I'll start by telling you the various kinds of differential forms, and the basic operations on them.
- In $n$-dimensional space, there are $n+1$ kinds of differential forms, from 0-forms to $n$-forms. You can think of a $k$-form as a $k$-dimensional density. A 0-form is a function, and a 1-form is a row-vector field (in coordinate-free language, a dual-vector field).
- The exterior derivative is an operation $d$ that turns $k$-forms into $(k + 1)$-forms. As its name suggests, it generalizes the operation of differentiating a function.
- The Hodge star is an operation $\star$ that turns $k$-forms in to $(n - k)$-forms. It comes from the dot product between column vectors. In fact, the Hodge star encodes the same geometric information as the dot product: if you know one, you can reconstruct the other.
- The codifferential is an operation $d^\dagger$ that turns $k$-forms into $(k - 1)$-forms. In an odd-dimensional space, like ordinary 3-dimensional space, applying $d^\dagger$ to a $k$-form is the same as applying $(-1)^k \star d \star$. In an even-dimensional space, $d^\dagger$ always acts like $-\star d \star$.
If you keep in mind that a 0-form is a function and a 1-form is a row-vector field, all the familiar operations of vector calculus can be written in terms of the ones above.
- The gradient of a function $f$ is the 1-form $df$.
- The curl of a 1-form $A$ is the 1-form $\star dA$.
- The divergence of a 1-form $A$ is the function $\star d \star A$.
- The Laplacian of a function or 1-form $\omega$ is $-\Delta \omega$, where $\Delta = dd^\dagger + d^\dagger d$. The operator $\Delta$ is often called the Laplace-Beltrami operator.
With this glossary in hand, you should be able to follow the steps of the calculation above, which is mostly just translating back and forth between languages. The only tricky bit is getting the sign right when you rewrite $d^\dagger$ as $\pm \star d \star$: you have to figure out what kind of form $d^\dagger$ is being applied to.
Best Answer
What I have so far is:
We look at the number next to the $\delta $, we then just have to work out the curl by doing the determinant matrix of our $\vec{F}$ and our partial derivatives of ${x,y,z}$, we get an answer for this in this the form ${a\vec{i},b\vec{j},c\vec{k}}$. Then we work out the curl of this equation ${a\vec{i},b\vec{j},c\vec{k}}$. We then plug our point in that corresponds to the $\delta$ value.