There exist $F$ for which there is no global solution $f$ to the above equation. Here is how you can construct an example:

First, regard $F$ as a vector field on $\mathbb{R}^3$ and consider its dual $1$-form $\phi$. The left hand side of your equation can then be written as the Hodge dual of $\phi\wedge\mathrm{d}\phi$ and the right hand side can be written as the Hodge dual of $\mathrm{d}f\wedge\mathrm{d}\phi$, so your equation becomes $$ \phi\wedge\mathrm{d}\phi = \mathrm{d}f\wedge\mathrm{d}\phi = \mathrm{d}\bigl(f\,\mathrm{d}\phi\bigr).\tag1 $$

Now suppose that $\phi$ has compact support and that the integral of $\phi\wedge\mathrm{d}\phi$ over $\mathbb{R}^3$ is nonzero. (See below for a construction of such a $\phi$.) Then integrating the ends of (1) over $\mathbb{R}^3$ and using Stokes' Theorem will yield a contradiction.

Now, to construct such a $\phi$, let $x,y,z$ be standard coordinates on $\mathbb{R}^3$ and let $h\ge0$ be a smooth function of compact support on $\mathbb{R}^3$ that is nonzero somewhere. Then set $$ \phi = h\,(\mathrm{d}z - y\,\mathrm{d}x).\tag2 $$ Then computation shows that $\phi\wedge\mathrm{d}\phi = h^2\,\mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z$ has compact support and its integral over $\mathbb{R}^3$ is positive, as desired.

There exist $F$ for which there is no global solution $f$ to the above equation. Here is how you can construct an example:

First, regard $F$ as a vector field on $\mathbb{R}^3$ and consider its dual $1$-form $\phi$. The left hand side of your equation can then be written as the Hodge dual of $\phi\wedge\mathrm{d}\phi$ and the right hand side can be written as the Hodge dual of $\mathrm{d}f\wedge\mathrm{d}\phi$, so your equation becomes $$ \phi\wedge\mathrm{d}\phi = \mathrm{d}f\wedge\mathrm{d}\phi = \mathrm{d}\bigl(f\,\mathrm{d}\phi\bigr).\tag1 $$

Now suppose that $\phi$ has compact support and that the integral of $\phi\wedge\mathrm{d}\phi$ over $\mathbb{R}^3$ is nonzero. (See below for a construction of such a $\phi$.) Then integrating the ends of (1) over $\mathbb{R}^3$ and using Stokes' Theorem will yield a contradiction.

Now, to construct such a $\phi$, let $x,y,z$ be standard coordinates on $\mathbb{R}^3$, and let $h\not\equiv0$ be a smooth function with compact support on $\mathbb{R}^3$. Set $$ \phi = h\,(\mathrm{d}z - y\,\mathrm{d}x).\tag2 $$ Then computation shows that $\phi\wedge\mathrm{d}\phi = h^2\,\mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z$. Hence, its integral over $\mathbb{R}^3$ is positive, as desired.

There exist $F$ for which there is no global solution $f$ to the above equation. Here is how you can construct an example:

First, regard $F$ as a vector field on $\mathbb{R}^3$ and consider its dual $1$-form $\phi$. The left hand side of your equation can then be written as the Hodge dual of $\phi\wedge\mathrm{d}\phi$ and the right hand side can be written as the Hodge dual of $\mathrm{d}f\wedge\mathrm{d}\phi$, so your equation becomes $$ \phi\wedge\mathrm{d}\phi = \mathrm{d}f\wedge\mathrm{d}\phi = \mathrm{d}\bigl(f\,\mathrm{d}\phi\bigr).\tag1 $$

Now suppose that $\phi$ has compact support and that its integral over $\mathbb{R}^3$ is nonzero. (See below for a construction of such a $\phi$.) Then integrating both sides of (1) over $\mathbb{R}^3$ and using Stokes' Theorem will yield a contradiction.

Now, to construct such a $\phi$, let $x,y,z$ be standard coordinates on $\mathbb{R}^3$ and let $h\ge0$ be a smooth function of compact support on $\mathbb{R}^3$ that is nonzero somewhere. Then set $$ \phi = h\,(\mathrm{d}z - y\,\mathrm{d}x).\tag2 $$ Then computation shows that $\phi\wedge\mathrm{d}\phi = h^2\,\mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z$ has compact support and its integral over $\mathbb{R}^3$ is positive, as desired.

There exist $F$ for which there is no global solution $f$ to the above equation. Here is how you can construct an example:

First, regard $F$ as a vector field on $\mathbb{R}^3$ and consider its dual $1$-form $\phi$. The left hand side of your equation can then be written as the Hodge dual of $\phi\wedge\mathrm{d}\phi$ and the right hand side can be written as the Hodge dual of $\mathrm{d}f\wedge\mathrm{d}\phi$, so your equation becomes $$ \phi\wedge\mathrm{d}\phi = \mathrm{d}f\wedge\mathrm{d}\phi = \mathrm{d}\bigl(f\,\mathrm{d}\phi\bigr).\tag1 $$

Now suppose that $\phi$ has compact support and that the integral of $\phi\wedge\mathrm{d}\phi$ over $\mathbb{R}^3$ is nonzero. (See below for a construction of such a $\phi$.) Then integrating the ends of (1) over $\mathbb{R}^3$ and using Stokes' Theorem will yield a contradiction.

Now, to construct such a $\phi$, let $x,y,z$ be standard coordinates on $\mathbb{R}^3$ and let $h\ge0$ be a smooth function of compact support on $\mathbb{R}^3$ that is nonzero somewhere. Then set $$ \phi = h\,(\mathrm{d}z - y\,\mathrm{d}x).\tag2 $$ Then computation shows that $\phi\wedge\mathrm{d}\phi = h^2\,\mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z$ has compact support and its integral over $\mathbb{R}^3$ is positive, as desired.