Deriving the vector field from a curl If $B= (x/r^3,y/r^3,z/r^3)$ find a vector $A$ such that $curl A = B$.
 My attempt:
$∂A_3∂Y−∂A_2∂Z=x/r^3$
$∂A_1∂Z−∂A_2∂y=y/r^3$
$∂A_2∂x−∂A_1∂y=z/r^3$
I assumed $A_3=0$, ending up having to solve:
$A_2=\int xr^{-3}dz$
Here, I got stuck. Is there an easier way of doing this?

I assume your book asks for any solution even though physically meaningless. Take for instance $A_3 = 0$. Then
\begin{equation}
B_1 = \frac{\partial A_3}{\partial y}-\frac{\partial A_2}{\partial z} = \frac{x}{(\sqrt{x^2+y^2+z^2})^3}
\end{equation}
Thus,
\begin{equation}
A_2 = -\int \frac{x}{(\sqrt{x^2+y^2+z^2})^3} \mathrm{d}z = \frac{-xz}{(x^2+y^2)\sqrt{x^2+y^2+z^2}}
\end{equation}
Similarly,
\begin{equation}
B_2 = \frac{\partial A_1}{\partial z}-\frac{\partial A_3}{\partial x} = \frac{y}{(\sqrt{x^2+y^2+z^2})^3}
\end{equation}
Thus,
\begin{equation}
A_1 = \int \frac{y}{(\sqrt{x^2+y^2+z^2})^3} \mathrm{d}z = \frac{yz}{(x^2+y^2)\sqrt{x^2+y^2+z^2}}
\end{equation}

Solving $\nabla \times \mathbf{A} = \mathbf{B}$ is a coupled partial differential equation, which is not complete unless boundary conditions are given to make solution $\mathbf{A}$ unique. Otherwise, the following answer assumes $\mathbf{A}$ is zero on boundary (decays at infinity).
By Helmholtz decomposition, every vector field $\mathbf{B}$ is decomposed to a curl-free and divergence-free component, by
\begin{equation}
\mathbf{B} = \nabla \times \mathbf{A} - \nabla \phi,
\end{equation}
where $\phi$ is the potential (the curl-free part, since $\nabla \times (\nabla \phi) = 0$), and $\mathbf{A}$ is vector potential (the divergence-free part, since $\nabla \cdot (\nabla \times \mathbf{A}) = 0$). Here
\begin{equation}
\mathbf{B} = \mathbf{0} - \nabla \left(\frac{1}{\sqrt{x^2 + y^2 + z^2}} \right) = -\nabla \left(\frac{1}{r}\right).
\end{equation}
It means $\mathbf{B}$ can be decomposed into a potential field $\phi=r^{-1}$ and no vector potential $\mathbf{A}$.
The potential is unique upto a constant, $\phi = \frac{1}{r}+c$ for any constant $c$. The vector potential is unique up to a gradient field, $\mathbf{A} = \mathbf{0} + \nabla \psi$ for any scalar field $\psi$.
Indeed, $\mathbf{B} = \frac{1}{r^2}\mathbf{e}_r$ is the Coulomb force, which is curl free and represented by potential above.

