= Solution
The wall position is $x_w(t)=\operatorname{Re}(ce^{-i\omega t})$, so its <velocity> is $\operatorname{Re}(-i\omega ce^{-i\omega t})$. The <kinematic boundary condition> equates the normal fluid <velocity> to the normal wall <velocity>. Linearizing the evaluation point to the mean wall position gives
$$
\boxed{\phi_x(0,z,t)=\operatorname{Re}(-i\omega ce^{-i\omega t}),\qquad f_x(0,z)=-i\omega c}.
$$
The other wall and bottom retain $f_x(L,z)=0$ and $f_z(x,-h)=0$, while the free surface gives $f_z(x,0)=(\omega^2/g)f(x,0)$.
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