The vector field $$\frac{\vec{A}\times\vec{x}}{(\vec{A}\times\vec{x})^2}$$ has zero curl, thus a scalar field $\varphi$ exists such that $$\nabla\varphi=\frac{\vec{A}\times\vec{x}}{(\vec{A}\times\vec{x})^2}$$ here $\vec{A}$ is constant and $\vec{x}$ is the position vector. It also turns out that $\nabla \cdot \dfrac{\vec{A}\times\vec{x}}{(\vec{A}\times\vec{x})^2} = 0$, the divergence is zero. Then $\varphi$ is harmonic, $$\Delta \varphi=0$$
How do I find $\varphi$? Tried to write the 3 PDEs in Maple but did not succeed to find a solution this way.
The curl is not zero at the origin. Choose a coordinate system in which the $z$-axis aligns with $\vec{A}$. Let $\gamma$ be a circle of radius $\epsilon$ in the $xy$-plane and $D$ the disk enclosed by $\gamma$ in $xy$ plane. Then Stokes theorem gives us $$ \int_D \nabla\times \left(\frac{\vec{A}\times \vec{x}}{|\vec{A}\times \vec{x}|^2}\right)\cdot d\vec{S}= \oint_\gamma \frac{\vec{A}\times \vec{x}}{|\vec{A}\times \vec{x}|^2}\cdot d\vec{x}=\frac{1}{|\vec{A}|\epsilon}\int_{0}^{2\pi} \epsilon d\theta =\frac{2\pi}{|\vec{A}|} $$ meaning the curl cannot be zero everywhere. So what is going on here? Let $\vec{A}=A\hat{z}$, then
$$ \vec{V}=\frac{\vec{A}\times \vec{x}}{|\vec{A}\times \vec{x}|^2}=\frac{1}{A}\frac{-y\hat{x}+x\hat{y}}{x^2+y^2}=\frac{1}{A}\frac{\hat{\phi}}{\rho} $$ where $(\rho, \phi,z)$ is the cylindrical coordinates. As you can see $\vec{V}$ is ill-defined on the $z$-axis (where $\rho=0$ and $\hat{\phi}$ has no meaning), so the domain of definition of $\vec{V}$ is in fact $\mathbb{R}^3-(\text{line tangent to }\vec{A})$. But this space is not simply connected, so $\nabla\times \vec{F}=0$ does not necessarily imply $\vec{F}=\nabla \varphi$.
However suppose we still want to naively find a function $\varphi$ such that $\vec{V}=\nabla\varphi$. In cyclindrical coordinates we have $$ \begin{aligned} \nabla f&=\frac{\partial f}{\partial \rho}\hat{\rho}+\frac{1}{\rho}\frac{\partial f}{\partial \phi}\hat{\phi}+ \frac{\partial f}{\partial z}\hat{z}\\ \nabla\cdot \vec{F}&=\frac{1}{\rho}\frac{\partial (\rho F_\rho)}{\partial \rho}+\frac{1}{\rho}\frac{\partial F_\phi}{\partial \phi}+ \frac{\partial F_z}{\partial z}\\ \nabla \times \vec{F}&=\left(\frac{1}{\rho}\frac{\partial F_z}{\partial \phi}-\frac{\partial F_\phi}{\partial z}\right)\hat{\rho}+\left(\frac{\partial F_\rho}{\partial z}-\frac{\partial F_z}{\partial \rho}\right)\hat{\phi}+ \frac{1}{\rho}\left(\frac{\partial (\rho F_\phi)}{\partial \rho}-\frac{\partial F_\rho}{\partial \phi}\right)\hat{z}\ \end{aligned} $$ So your vector field is in fact equal to (naively) $$ \vec{V}= \nabla \left(\frac{\phi}{A}\right) $$ Can you see the problem now? Whatever $\phi$ is, it is not a single-values function. As you go one-times around the origin $\phi$ picks up an extra $2\pi$. The curl of $\vec{V}$ if you still want to stay faithful to this naive picture, should be interpreted as $\nabla \times \vec{V}=\frac{2\pi\hat{z}}{A}\int_{-\infty}^\infty dt\delta^3(x\hat{x}+y\hat{y}+(t-z)\hat{z})$ where $\delta^3(\vec{x})$ is the 2-D Dirac delta function vanishing everywhere but the origin [Hopefully I haven't done any mistakes in determining this curl]. Physicists use these (naive) tricks all the time by the way. In their language, the origin is an (infinitesimal) vortex, and this vector field can be thought of circulating flow field around the vortex.