Dispersion equation with variable wavenumber The wave equation
$$u_{tt}=c^2 u_{xx}$$ is known to have a simple wave solution $u(x,t)=Ae^{i(kx-\omega t)}$ where the dispersion equation is simply $c=\omega/k$. Yet, let the wavenumber be a function in $x$, then the independent variable $x$ will appear in the dispersion solution cause the first and second derivatives are functions in $x$ as the following:
$$ \dfrac{\partial{u}}{\partial x} = (ik+ixk_x) e^{i(kx-\omega t)}$$ and $$ \dfrac{\partial^2 u}{\partial x^2} =  \left( (ik_x+ik_x+ixk_xx) + (ik+ixk_x)^2 \right) e^{i(kx-\omega t)} .$$
then $$c^2=\dfrac{-\omega^2}{(2ik_x+ixk_xx)+(ik+ixk_x)^2}$$
Did anyone encounter an independent variable as $x$ explicitly in the dispersion relationship as in the terms $ixk_x x$ and $ixk_x$?
 A: Effectively, $u(x,t) = A e^{i(kx-\omega t}$ (with $c= \omega/k$) is a solution of the equation. The problem is in supposing that:
$$\tilde{u}(x,t) = A e^{i(k(x)x-\omega t}$$
is also a solution of the same equation, that step does not seem justified, in fact, for almost any choice of $k(x)$ we will not have a solution in the strict sense. The most general solution formed by plane waves would be a generalized linear combination of the type:
$$u(x,t) = \int_{-\infty}^{+\infty} A(k) e^{ik(x - ct)}\ \text{d}k$$
Given some function $k(x)$ there is no guarantee that you could found a function $A(k)$ such that:
$$e^{i(k(x)x-\omega t)} = \int_{-\infty}^{+\infty} A(k) e^{ik(x - ct)}\ \text{d}k$$
A: 
Is it normal to encounter...

It is difficult to answer "is it normal?" Normal to whom? That being said, in my opinion, this is not normal/typical.

and what does it mean?

No one can answer "what does it mean?"
No one can answer because you have posited some arbitrary function of $x$ that you just happen to be calling "$k$". No one knows what this arbitrary function is, so no one can tell you what it "means."

In the comments I suggested a more complete example (with more context) wherein:
$$
k(x)=\sqrt{2m(E-V(x))/\hbar^2}\;,
$$
where $E$ is a particle's energy, and $V(x)$ is the potential it sees.
With this additional context, I can, for example, point you to the WKB approximation, where you can try to glean some physical meaning.
But, if you just have some arbitrary function:
$$
k(x) = ??????
$$
no one can provide a physical meaning.

Update: OP has edited their post to remove the two above-quoted questions. So I will update my answer...
OP's new question is:

Did anyone encounter an independent variable as $x$ explicitly in the dispersion relationship as in the terms $ixk_x x$ and $ixk_x$?

You will likely not encounter such a spatial dependent "dispersion relationship" unless you are trying to force plane waves to be a solution to a problem not solved by plane waves.
All one can do is provide examples. Above, I provided the example of the WKB approximation.
Another example to consider is the propagation of waves through bulk materials with an interface. This is similar to the one-dimensional problem of a wave travelling along a rope whose linear density $\mu$ depends on position. Suppose, for example, the rope is made up of one heavy rope (with linear density $\mu_1$) and one light rope (with linear density $\mu_2$) joined at a knot. Far from the knot the dispersion relation is either:
$$
\omega = k\sqrt{\frac{T}{\mu_1}}
$$
or
$$
\omega = k\sqrt{\frac{T}{\mu_2}}\;,
$$
which could be though of as a spatially dependent dispersion relation.
As discussed above, I can provide some examples and discuss their meaning. But, given that OP's $k(x)$ is some unknown/arbitrary function, it is not possible to discuss a particular physical meaning.
A: From a mathematical point of view your idea does not make sense. Remember that plane waves do not exist, they are just a tool that physicists introduce to build up a mental representation. Ask a mathematician; he will tell you that to get a dispersion relation, you just need to take the Fourier transform of your wave equation. But (x,k) is the pair of variables involved in the Fourier transform. They can only appear together in: $ e^{i(k.x-\omega t)}$. with k being the Fourier variable independent of x.
A: I came across similar solutions in studying hydrodynamic (in)stability.
These solutions, usually in the form of traveling waves $\phi(x,t) = A e^{i(k(\omega)x - w t)}$, appear in the study of weakly nonlinear instabilities in open flows as the solution of weakly nonlinear approximation of the equations of motion, and not as the results of the linear wave equation.
Now, I can't really remember if I've ever seen the dispersion you wrote in your answer, but I'll leave a list of references here where you can have a look by yourself:

*

*Huerre, Rossi Hydrodynamic instabilities in open flows, 1998.
(not easy to found, if you have no access to university library, and maybe even if you have it)

*Charru, Hydrodynamic Instabilities

*Drazin, Hydrodynamic Stability
You can also look for Landau-Ginzburg equation, that usually results from the weakly nonlinear approximation of nonlinear equations of motion in fluid mechanics, and other fields, and it's a mathematical model often used for introducing and exploring weakly nonlinear instabilities.
If you give us some details about your field of study, maybe someone could give you more precise suggestions.
EDIT: Wave function with the desired solution
It seems to me that the function $\phi(x,t) = A e^{i(k(x)x - \omega t)}$ is the solution of a the following PDE with non uniform coefficients, that goes into the wave equations in uniform media when coefficients are uniform, reading
$\dfrac{\partial^2 \phi}{\partial t^2} = \dfrac{\omega^2}{\kappa(x)'} \dfrac{\partial}{\partial x} \left( \dfrac{1}{\kappa(x)'}\dfrac{\partial \phi}{\partial x} \right) = \dfrac{\omega^2}{\kappa(x)'^2} \dfrac{\partial^2 \phi}{\partial x^2} - \dfrac{\omega^2 \kappa(x)''}{\kappa(x)'^3}  \dfrac{\partial \phi}{\partial x}$,
having defined $\kappa(x) := k(x) x$, and their derivatives
$\kappa(x)' = k(x) + k(x)'x$,
$\kappa(x)'' = 2 k(x)' + k(x)'' x$.
EDIT 2: performing a change of variable $y(x)$ so that $y'(x) = k'(x)/ k_0$, and $\partial/\partial y = k_0 / k'(x) \, \partial/\partial x$ you get the classical linear wave equation in $y$, $t$ independent variables.
Maybe this can help the imagination or the memory of someone, to think at an example where this kind of equation is used.
A: Since you have only one wave $k=\frac{2\pi}{\lambda}=cst$, so $k_{x}=0$ ,  your formula is simpler : $$c^{2}=\frac{-\omega^{2}}{(ik)^{2}}$$
i.e.$$k=\frac{\omega}{c}$$
Note: the last relation gives:
$$(2+x^{2})k_{x}i=k+xk_{x}-\frac{\omega^{2}}{c^{2}}$$
$k(x)\in \mathbb{R}$, for the equation to be homogeneous,
in the left side of the equation  $k_{x}(x)$ must be pure imaginary, CONTRADICTION.
PS: The only theory i know that takes into account the variation of the frequency,length, vector,...., of wave with the variable space (height x) is general relativity: Einstein's shift https://en.wikipedia.org/wiki/Gravitational_redshift
