Directional derivative doubt I have two functions.
$e(t,z)=\cos(\omega t)\cos(\beta z)$, $h(t,z)=\frac{1}{\eta}\sin(\omega t)\sin(\beta z)$
Let's set the direction $\hat{u} = \hat{z} - c\hat{t}$, with $\dfrac{\epsilon_0}{c}=\dfrac{1}{\eta}$ and $\dfrac{\omega}{c}=\beta$. I can calculate $\dfrac{\partial( h-\frac{1}{\eta} e)}{\partial u}$ and I find it to be 0. So the quantity $h-\dfrac{1}{\eta} e$ is constant along that direction and since I know that for some $(t,z)$ points both $e$ and $h$ are $0$, that constant is $0$. So I'd like to say that if I change $z$ and $t$ according the $u$ direction, the ratio $\dfrac{e}{h}$ stays constant. It seems to me quite wrong but I'd like to get some comments about this misunderstanding.

There seems to be some confusion regarding directional derivatives vs. partial derivates. 


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*Your function $f(t,z)=h-\frac{1}{\eta} e= -\frac{1}{\eta} \cos(\beta(z+ct))$ is constant in the direction $\hat{u}=(-\frac{1}{c},1)$ in the $(t,z)$-plane, since:
$$ -\frac{1}{c} \partial_t f + \partial_z f = 0$$ 

*If you make a change of variables, say $u=z-ct$, $v=z+ct$, your function becomes $\hat{f}(u,v)=-\frac{1}{\eta} \cos (\beta v)$ which is independent of $u$, whence the partial derivative w.r.t. u is zero of this function in the coordiantes $(u,v)$. It seems that you are mixing the two notions.
If you happen to know that $f(t_0,z_0)=0$ then along the $\hat{u}$ direction, 
$f(t_0-\frac{1}{c}s, z_0+s)=0$ for all $s\in {\Bbb R}$ implying that $h=\frac{1}{\eta} e$ along that line, so the ratio $e/h=\eta$ whenever defined (note that sometimes both vanish).
Explicitly, we have that $\beta(z+ct)=\beta z + w t$ is constant equal to $\pi/2+2\pi k$ for some integer $k$ and since $\;\cos(\pi/2-x)=\sin(x)\;$ we get  $e(t,z)= \cos(wt) \sin(wt) = \eta h(t,z)$ along that line.
