How does a drum's membrane oscillate? I was thinking about a problem with  a simple wooden drum , constructed but just a membrane and a wooden circular part .Let's say the mebrane is streched by a constant tension $T$ and it has a density $\rho$ We know that the points of the perimeter are fixed .
The motion of the membrane can be described by: $T(\frac{d^2z}{d^2x} + \frac{d^2z}{d^2y})=\rho\frac{d^2z}{d^2t}$ So if we are looking for the normal modes of the mebrane we can assume that : the solutions have the form : $z(x,y,t) = Acos(k_xx + \theta_x)cos(k_yy + \theta_y)cos(\omega t + \phi)$

Where I get stucked is whether every point of the surface remains fixed in z axis ( there is no motion up and down) and all that happens is:" when we hit the mebrane with the baguette we create a travelling wave that gets reflected when it reaches the finite endings of the membrane and in the pernament situation there are surface standing waves" or if the mebrane actually oscillates with $A \to 0$ but still $A \neq 0 $ in the z axis ...
Any ideas?
 A: Beware: just because you can plug your equation into the differential equation and it works does not mean it is a valid solution for your physical system. You are forgetting about boundary conditions.
Your proposed equation is not a solution because the drum is a circular membrane. This means you need to satisfy the boundary condition $z(x,y,t)=0$ at all times whenever the point $(x,y)$ lies on the edge of the drum.
Looking at your proposed "solution", $z=0$ when $x=(1/k_x)\cdot(m\pi/2-\theta_x)$ or when $y=(1/k_y)\cdot(n\pi/2-\theta_y)$ for odd integers $m$ and $n$. The collection of these points describes a regularly spaced grid, not a circle.$^*$
You need to determine solutions that are valid with the appropriate boundary conditions. Fortunately, this has been done many times before. The typical approach is to convert your coordinate system into polar coordinates, so that $z=0$ when $r=R$, the radius of the drum. Then you can show that solutions can be expressed as linear combinations of Bessel functions, and the linear combination depends on what your initial conditions of the membrane are.
More detail can be found in the above links, or probably any decent text book on partial differential equations, mathematics for the physical sciences, etc.

$^*$ So, your solution could be valid for vibrations on a rectangular membrane.
A: Firstly, your differential equation is a partial differential equation, so it needs partial derivatives.
Because the drum is circular you have to transform your first equation into polar coordinates:
$$T\Big(\frac{1}{r}\frac{\partial}{\partial r}\Big(r\frac{\partial z}{\partial r}\Big)+\frac{1}{r^2}\frac{\partial^2z}{\partial \varphi^2}\Big)=\rho\frac{\partial z^2}{\partial t^2}$$
So we're looking for a function:
$$z(r,\varphi,t)$$
Reworked and with PDE shorthand:
$$z_{rr}+\frac{1}{r}z_r+\frac{1}{r^2}z_{\varphi \varphi}=\frac{1}{c^2}z_{tt}\tag{1}$$
with:
$$c^2=\frac{T}{\rho}$$
We also have a boundary condition:
$$z(R,\varphi,t)=0\tag{2}$$
where $r=R$ is the radius of the drum.
(Later we'll find another boundary condition, as two are required for a second order PDE, for each variable)
And an initial condition:
$$z(r,\varphi,0)=f(r,\varphi)$$

Because of the homogeneous boundary condition $(2)$ this PDE is probably solvable with [Separation of Variables][1]. Because this is a HW&E question I will only point you in the right direction.
As starting assumption ('Ansatz') we use:
$$z(r,\varphi,t)=R(r)\Phi(\varphi)T(t)$$
Or for short:
$$z=R\Phi T$$
Now insert the Ansatz into the PDE $(1)$:
$$\Phi TR''+\frac{1}{r}\Phi TR'+\frac{1}{r^2}RT\Phi''=\frac{1}{c^2}R\Phi T''$$
Divide everything by $R\Phi T$:
$$\frac{R''}{R}+\frac{R'}{rR}+\frac{\Phi''}{r^2 \Phi}=\frac{1}{c^2}\frac{T''}{T}$$
Notice how we've separated $T$ form both other functions, so we can write:
$$\frac{R''}{R}+\frac{R'}{rR}+\frac{\Phi''}{r^2 \Phi}=\frac{1}{c^2}\frac{T''}{T}=-m^2$$
where $m^2$ is a Real number known as a separation constant.
We now have $2$ equations:
$$\frac{1}{c^2}\frac{T''}{T}=-m^2\tag{3}$$
and:
$$\frac{R''}{R}+\frac{R'}{rR}+\frac{\Phi''}{r^2 \Phi}=-m^2$$
Multiply with $r^2$:
$$\frac{r^2R''}{R}+\frac{rR'}{R}+\frac{\Phi''}{\Phi}=-m^2 r^2$$
$$\frac{r^2R''}{R}+\frac{rR'}{R}+m^2 r^2=-\frac{\Phi''}{\Phi}=-l^2$$
Which gives us two more equations:
$$-\frac{\Phi''}{\Phi}=-l^2\tag{4}$$
$$\frac{r^2R''}{R}+\frac{rR'}{R}+m^2 r^2+l^2=0\tag{5}$$
And this is where the fun starts for you: $(3)$, $(4)$ and $(5)$ are three ordinary differential equations for you to solve!
1 https://en.wikipedia.org/wiki/Separation_of_variables
