The first thing to check isIf you switch the symbolsecond and the characteristic variety (where the symbol is degenerate). Here, the symbol is $\sigma = A_1\xi_1 + A_2\xi_2 + A_3\xi_3$, where $A_1$, $A_2$, $A_3$ are the matrix coefficientsthird rows of $\partial_x$, $\partial_y$, and $\partial_z$. For fixed $(x,y,z)$your system, the characteristic varietydifferential operator is given by the equation $$ 0 = \det \sigma = f_1f_2f_3\xi_1\xi_2\xi_3. $$ Therefore, assuming $f_1, f_2, f_3$ are all nonzero, the system is nondegenerate, andsame as the characteristic variety consists of 3 transversal planes. This is preserved under changelinearization of variablesequation (4. In particular, if3) in Existence of elastic deformations with prescribed principal strains and triply orthogonal systems at a point where \begin{align*} u &= x+y+z\\ v &= y\\ w &= z, \end{align*}$$ g_{\alpha\beta} = \delta_{\alpha\beta} \text{ and }\frac{\partial x^\alpha}{\partial y^i} = \delta_{\alpha i}. $$ thenAs mentioned immediately after the equation becomes $$ (A_1+A_2+A_3)\partial_u N + A_2\partial_vN + A_3\partial_wN = \omega. $$ Since $A_1+A_2+A_3$, this system is invertiblenot (if $f_1,f_2,f_3$ are nonzerosymmetric), you hyperbolic. It also can multiply both sides by its inverse and get an equation of the form $$ \partial_u N + B\partial_vN + C\partial_wN = \hat\omega. $$ The symbol ofnever be elliptic. It follows that this system looks like $$ \hat\sigma = I\hat\xi_1 + B\hat\xi_2 + C\hat\xi_3. $$ Since the characteristic variety still consists of 3 distinct planes, this implies that $\hat\sigma$ is smoothly diagonalizabledifficult to solve. Therefore, the system is a strictly hyperbolic system, and the initial value problem, where $N$ is prescribed on the plane $u = 0$, is well-posed and therefore has a solution on allI don't know of $\mathbb{R}^3$, if $f_1, f_2, f_3$ never vanishany work in this direction.
In factthe paper cited, by using other linear changeswe observe that the obstruction is due to a type of variables,gauge invariance and reformulate the initial value problem with $N$ prescribed on any $2$-plane, where $\xi_1\xi_2\xi_3\ne 0$in a way that eliminates this issue, is well-posedresulting in a symmetric hyperbolic system that can be solved.
If $f_1f_2f_3 = 0$ somewhere, then it is possible, with some work piecing together solutions of different initial value problems, to identify the largest domain on which there is a solution.Perhaps your problem can be treated similarly?