coefficient of bivariate generating functions how can I find/what is the coefficient of the $x^4y^4$ and $x^6y^6$ term for the following generating function
$$\frac{1}{1-x-y-x^2y}$$
I've been informed I can do this by using Pascal's triangle, by adding an extra value of 2 back and 1 down but am unsure of how to? (I'm new to combinatorics)
 A: A method to find the coefficient of $x^m y^n$ in $\frac{1}{1-x-y-x^2y}$ is described as follows.
First, let
$$\frac{1}{1 - x - y - x^2y} = a_0(x) + a_1(x) y + a_2(x) y^2 + a_3(x) y^3 + \cdots .$$
We need to determine $a_n(x) y^n$.
Clearly,
$$\frac{\partial^n  \tfrac{1}{1 - x - y - x^2y}}{\partial y^n} \Big\vert_{y=0} = n! a_n(x).$$
It is easy to obtain (since $1 - x - y - x^2y$ is affine in $y$)
$$\frac{\partial^n  }{\partial y^n} \frac{1}{1 - x - y - x^2y}
= \frac{(-1)^n n! (-1-x^2)^n}{(1-x-y-x^2y)^{n+1}}.$$
Thus, we have
$$a_n(x) = \frac{(1+x^2)^n}{(1-x)^{n+1}}.$$
By noting that $\frac{1}{1-x} = \sum_{j=0}^\infty x^j$ and $\frac{\partial^n }{\partial x^n} \frac{1}{1-x} = \frac{n!}{(1-x)^{n+1}}$, we have
$$\frac{1}{(1-x)^{n+1}} = \frac{1}{n!}\frac{\partial^n }{\partial x^n}\sum_{j=0}^\infty x^j = \sum_{j=0}^\infty {n+j \choose j} x^j.$$
Also, 
$$(1+x^2)^n = \sum_{k=0}^n {n\choose k} x^{2k}.$$
Thus, we have
\begin{align}
a_n(x) &= \sum_{k=0}^n {n\choose k} x^{2k} \cdot \sum_{j=0}^\infty {n+j \choose j} x^j\\
&= \sum_{k=0}^n \sum_{j=0}^\infty {n\choose k} {n+j \choose j} x^{2k+j}.
\end{align}
The coefficient of $x^m$ in $a_n(x)$ is given by
\begin{align}
a_{mn} &= \sum_{2k+j = m, \ j\ge 0, \ 0\le k \le n} {n\choose k} {n+j \choose j} \\
&= \sum_{k=0}^{\min(n, \lfloor \frac{m}{2}\rfloor)} {n\choose k} {n+m-2k \choose m-2k}. 
\end{align}
For example, 
$$a_{44} = \sum_{k=0}^2 {4\choose k} {8-2k \choose 4-2k} = 136$$
and
$$a_{66} = \sum_{k=0}^3 {6\choose k} {12-2k \choose 6-2k} = 2624.$$
Remark: In general, let
$$f(x, y) = a_{00} + a_{10}x + a_{01}y + a_{20}x^2 + a_{11}xy + a_{02}y^2 
+ \cdots.$$
Clearly, 
$$
\frac{\partial^{m+n} f(x,y) }{\partial x^m \partial y^n}\Big\vert_{(x,y)=(0,0)}
= m! n! a_{mn}$$
which results in
\begin{align}
a_{mn} = \frac{1}{m!}\frac{1}{n!}
\frac{\partial^m }{\partial x^m}
\Big(\frac{\partial^n  f(x,y)}{\partial y^n} \Big\vert_{y=0}\Big)\Big\vert_{x=0}.
\end{align}
