Eigenvalues of an element in a Weyl algebra I have an operator acting on the polynomial algebra $\mathbb{C}[x,y,z]$ that I would like to find the eigenvalues/eigenvectors of.  More specifically, let $P(x_1, \ldots, x_6)$ be a homogeneous polynomial, my operator has the form $P(x,y,z, \frac{\partial}{\partial x},  \frac{\partial}{\partial y},  \frac{\partial}{\partial z})$.  Are there any general strategies that could help me?  For instance, say my operator were: $$z^2y\frac{\partial^3}{\partial x^2 \partial y} + y^3z\frac{\partial^4}{\partial y^2 \partial z^2} + xz^2\frac{\partial^3}{\partial x \partial z^2}.$$  My actual operator is degree 6 and more complicated, but other than that the same type of object. Any thoughts or references on how to attack this type of problem will be very welcome. Thanks a lot!
EDIT: My first example-operator was very poorly chosen, since every monomial would automatically be an eigenvector. I have now altered it a little to avoid this. Keep in mind this was only an example to show what type of object I am considering, and I am looking for general strategies. None the less, thanks for the quick response. 
EDIT 2: I had misread the degree of my actual operator - it is 6. 
 A: If the polynomial $P(x_1, \ldots, x_6)$ is homogeneous of degree zero (where $x_1, x_2, x_3$ are considered of degree 1 and $x_4, x_5, x_6$ of degree -1), the computation of the eigenvalues/eigenvectors reduces to linear algebra. Indeed, since the operator $P(x,y,z, \frac{\partial}{\partial x},  \frac{\partial}{\partial y},  \frac{\partial}{\partial z})$ preserves total degree, it is enough to consider each graded piece $\mathbb{C}[x,y,z]_d$ separately, and the latter are finite-dimensional vector spaces on which the operator acts linearly. Your explicit example is of this kind.
If $P$ is of homogeneous of negative degree, then it can only have zero as an eigenvalue, since the associated operator is nilpotent. For $P$ of positive degree it also seems clear that the only possible is zero.
A: NB: This answers an old version of the question...
The differential operator in your example is very, very special: each monomial in $x$, $y$ and $z$ is an eigenvector, so it is in fact diagonalizes is the basis of monomials of $k[x,y,z]$!
