The Delannoy path is a set of paths from (0, 0) to (m, n) using only steps (1, 0), (1, 1) or (0, 1). I know the generating function is $\frac{1}{1-x-y-xy}$
So far Im given, D(x, y) = $\sum_{m \geq 0}\sum_{n \geq 0} d_{m,n}x^{m}y^{n}$ and I know $d_{m,n}$ = D(m-1, n) + D(m-1, n-1), D(m, n-1).
What properties should I use to go from the recursion, to the generating function?
Start with
$\begin{array}\\ d(x, y) &=\sum_{m \geq 0}\sum_{n \geq 0} d(m,n)x^{m}y^{n}\\ &=-d(0, 0)+\sum_{n \geq 0} d(0,n)y^{n} +\sum_{m \geq 0} d(m,0)x^{m} +\sum_{m \geq 1}\sum_{n \geq 1} d(m,n)x^{m}y^{n}\\ &=-d(0, 0)+d_1(y)+d_2(x) +\sum_{m \geq 1}\sum_{n \geq 1} d(m,n)x^{m}y^{n}\\ &=-d(0, 0)+d_1(y)+d_2(x) +\sum_{m \geq 1}\sum_{n \geq 1} (d(m-1, n) + d(m-1, n-1)+d(m, n-1))x^{m}y^{n}\\ \end{array} $
Use the initial conditions to get $d_1(y)$ and $d_2(y)$.
Split the last sum into $\sum_{m \geq 1}\sum_{n \geq 1} d(m-1, n)x^{m}y^{n} +\sum_{m \geq 1}\sum_{n \geq 1} d(m-1, n-1)x^{m}y^{n} +\sum_{m \geq 1}\sum_{n \geq 1} d(m, n-1)x^{m}y^{n} $ and fiddle with the exponents of $x$ and $y$ to get each sum in terms of $x, y, d(x, y), d_1(y),$ and $d_2(x) $.
This will give you a formula for $d(x, y)$.