Let $\boldsymbol{\mathsf{H}}$ be the Hessian of the function $F(\boldsymbol{x})$. If this function is of the form $$ F(\boldsymbol{x}) = f(\hat{\boldsymbol{\omega}}\cdot\boldsymbol{x}) $$ with some unit vector $\hat{\boldsymbol{\omega}}$ and function $f$, then $$ \mathsf{H}_{ij} = \hat{\omega}_i\,\hat{\omega}_j\,f''(\hat{\boldsymbol{\omega}}\cdot\boldsymbol{x}). $$ In particular, $\boldsymbol{\mathsf{H}}$ has rank one everywhere with the only non-trivial eigenvector being $\hat{\boldsymbol{\omega}}$.
My question, is the reverse also true? That is, if $\boldsymbol{\mathsf{H}}$ has rank one everywhere, is then necessarily $F(\boldsymbol{x})$ of the form above?
No; take as a 2D example:
$$F(x, y) = \sqrt{x^2 + y^2}$$