It's maybe a silly question but I was wondering if there exists a closed 1-form $\alpha$ on the manifold $\mathbb{R}^3 -\{0\}$ of the form $$\frac{1}{x^2+y^2+z^2}(adx+bdy+cdz)$$ with $a, b$ and $c$ smooth function on $\mathbb{R}^3 - \{0\}$? (with $\alpha \neq 0$ and $\alpha \neq dx+dy+dz$ )
Extra: what if I also require that $d(adx+bdy+cdz) \neq 0$?
Since $\mathbb R^3\setminus \{0\}$ is simply-connected, a closed $1$-form must be exact; that is, of the form $df$ for some function $f$.
The requirement $\alpha = \frac{1}{x^2+y^2+z^2}(adx+bdy+cdz)$ with $a,b,c\in C^\infty(\mathbb R^3\setminus \{0\})$ is not saying anything other than $\alpha\in C^\infty(\mathbb R^3\setminus \{0\})$. But from the comments it looks like you don't want cancellation: that is, $a,b,c$ should not vanish at the origin.
I'll impose an even more restrictive condition: $a,b,c$ homogeneous of degree $0$, that is constant on every line through the origin. Then $\alpha$ is homogeneous of degree $-2$, hence $f$ must be homogeneous of degree $-1$. The first such function that comes to mind is $$f(x,y,z)=(x^2+y^2+z^2)^{-1/2}$$ with $$\alpha = df = -\frac{x\,dx+y\,dy+z\,dz}{(x^2+y^2+z^2)^{3/2}}$$