If you assume
$\displaystyle u(x,t)=f(x)e^{-i \omega t} \implies u_{tt}=(i\omega)^2f(x)e^{-i\omega t}$
and
$\displaystyle u_{x}= f'(x)e^{-i \omega t} \text{ and } u_{xx}=f''(x)e^{-i \omega t}$
Now just expand out the right hand side of the eqation
$\displaystyle \frac{\partial }{\partial x} \left( A(x) \frac{\partial u}{\partial x}\right) = \frac{\partial A}{\partial x}\frac{\partial u}{\partial x}+A(x)\frac{\partial^2 u}{\partial x^2}=A'u'+Au''$
Now just put all of this into the equation
$\displaystyle (i \omega)^2f(x)e^{-i \omega t} = \frac{c^2}{A}\left(A'f'e^{- i \omega t}+f''e^{-i \omega t} \right)$
Can you finish from here?