Consider a $d$-regular graph $G = (V, E)$ of order $n$ such that $|N_G(x) \cap N_G(y)| = \lambda$ for all distinct $x, y \in V$. By double counting we have a necessary condition $\lambda (n - 1) = d(d - 1)$. My question is how do I determine whether such a graph exists? Are there any sufficient conditions for that?
This question arises from a special case when $\lambda = 6$, $d = 15$ and $n = 36$. I don't know how to show the existence of this special case either (except to really construct such a graph).
You are looking for an $(n,d,\lambda, \lambda)$ strongly regular graph. The normal labelling of the parameters are $(v,k, \lambda, \mu)$, so you are considering the special case where $\lambda = \mu$.
There are lots of necessary conditions for the existence of a strongly regular graph. In particular, there are formulas to give the eigenvalues (there are always 3) and their multiplicities; conditions such as the Krein conditions rely on the fact that these multiplicities must be nonnegative integers. Of course even when these conditions hold, you aren't guaranteed that an example exists until you construct one. There are many parameter sets where existence is unknown.
For specific parameters, Brouwer maintains a nice list here. For your special case of $(v,k,\lambda, \mu) = (36,15,6,6)$, there are 32548 nonisomorphic examples.