There are two methods; one is to translate the information into matrix form and work with that. The other is to work with $L$ directly.
For the first method, the first step is to find a basis for your vector space. In this case, you can take the basis of all monomials of degree at most $n$.
Then apply $L$ to the vectors of the basis in order to find the coordinate matrix of $L$; then work with it.
For example, for $n=2$, a basis would be $\beta=\{1,x,x^2,y,xy, y^2\}$ (note the dimension is not $3$, as you conjectured, but in this case $6$; the dimension is equal to the number of monomials of degree at most $n$).
(Note: I'm interpreting your problem as being with polynomials of total degree at most $n$; if you meant that the degree in each of $x$ and $y$ is at most $n$, then for $n=2$ you would need to add $x^2y$, $xy^2$, and $x^2y^2$ to the basis, and include them in the computations below; the main idea is the same, though.)
Then you apply $L$ to each vector in the basis, and express the answer as a linear combination of the basis:
$$\begin{align*}
L(1) &= \frac{\partial}{\partial x}1 + \frac{\partial}{\partial y}1= 0 + 0\\
&=0\cdot 1 + 0 x + 0x^2 + 0y + 0xy + 0 y^2,\\
L(x) &= =\frac{\partial}{\partial x}x + \frac{\partial}{\partial y}0 = 1 + 0\\
&=1\cdot 1 + 0 x + 0x^2 + 0y + 0xy + 0 y^2,\\
L(x^2) &= \frac{\partial}{\partial x}x^2 + \frac{\partial}{\partial y}x^2 = 2x + 0\\
&= 0\cdot 1 +2x + 0x^2 + 0y + 0xy + 0y^2,\\
L(y) &= \frac{\partial}{\partial x}y + \frac{\partial}{\partial y}y = 0+1\\
&= 1\cdot 1 + 0x + 0x^2 + 0y + 0xy + 0y^2,\\
L(xy) &= \frac{\partial}{\partial x}xy + \frac{\partial}{\partial y}xy= y+x\\
&= 0\cdot 1 + 1 x + 0x^2 + 1y + 0xy + 0y^2,\\
L(y^2) &= \frac{\partial}{\partial x}y^2 + \frac{\partial}{\partial y}y^2 = 0 + 2y\\
&= 0\cdot 1 + 0x + 0x^2 + 2y + 0xy + 0y^2.
\end{align*}$$
Now, relative to the basis, the coordinate matrix of $L$ is:
$$\left(\begin{array}{cccccc}
0 & 1 & 0 & 1 & 0 & 0\\
0 & 0 & 2 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 1 & 2\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0
\end{array}\right)$$
Now that you have a matrix, it is an easy matter to find the characteristic polynomial (it is $t^6$), the eigenvalues (only $\lambda=0$); the eigenspace (there are three free variables, so the dimension of the eigenspace is $3$); the eigenvectors (spanned by $1$, $x-y$, and $x^2 - 2xy + y^2$), the the Jordan canonical form (check the dimension of the nullspaces of $L^2$ (it's dimension is $5$), and of $L^3$ (which is all of the space). This gives you the Jordan canonical form (three blocks total, largest block of size $3$, so it must be one $3\times 3$ block, one $2\times 2$ block, one $1\times 1$ block).
For the other method you would have to work directly with $L$ and polynomials Solve the equation $L(a+bx+cx^2+dy + exy+fy^2) = \alpha(a+bx+cx^2+dy+exy+fy^2)$ to find that the only solutions have $\alpha=0$ (giving you the eigenvalues), and finding the explicit solutions gives the eigenvectors. Note that $L^2$ is not the zero linear transformation, then that $L^3$ is; this gives all the information you need.
(You will find that in the general case something similar occurs: you cannot have any eigenvalues other than $0$, because $L$ always reduces the maximum degree of $p(x)$ by $1$; that you need $n$ applications of $L$ to get the zero linear transformation; and you should find a pretty simple formula for the dimension of the space (hence the size of the matrix) and a nice way of seeing the Jordan canonical form.)