Let $F=\frac{(x,y,z}{(a^2x^2+b^2y^2+c^2z^2)^{1.5}}$ be a vector field. Is there a vector field $G$ such that $curl(G)=F$?

Question

Let $A =\mathbb{R}^3 \setminus\{ (0,0,0)\},a,b,c>0$.

Let $F=\frac{(x,y,z}{(a^2x^2+b^2y^2+c^2z^2)^{1.5}}$ be a vector field.

Is there a vector field $G$ such that $curl(G)=F$ ?

I have no idea how to approach this problem , any help is welcome.