Solution:

Firstly, there are $\binom{36}{4}$ possible combinations of points. Call a square proper if its sides are parallel to the coordinate axes and improper otherwise. Note that every improper square can be inscribed in a unique proper square. Hence, an $n \times n$ proper square represents a total of $n$ squares: 1 proper and $n-1$ improper.

There are thus a total of
$$\begin{aligned} \sum_{i=1}^{6} i(6-i)^{2} & =\sum_{i=1}^{6}\left(i^{3}-12 i^{2}+36 i\right) \\ & =\sum_{i=1}^{6} i^{3}-12 \sum_{i=1}^{6} i^{2}+36 \sum_{i=1}^{6} i \\ & =441-12(91)+36(21) \\ & =441-1092+756 \\ & =105 \end{aligned}$$
squares on the grid. Our desired probability is thus $\frac{105}{\binom{36}{4}}=\frac{1}{561}$.