The number of nodes of each row of your tree is $2^d$ where d is the depth, and the first row is $d=0$. The number of nodes up to and including row $d$ is $2^{d+1}$. The inverse of $2^n$ is $\log_2 n$.
If you have node $n$ then its depth the smallest $d$ such that $2^{d+1} \le n$, which is $\lfloor \log_2 (n+1) \rfloor$. Using Python, your nodes 6 and 10 are at depths:
>>> import math
>>> math.floor(math.log2(6+1))
2
>>> math.floor(math.log2(10+1))
3
and more programmatically:
>>> for i in range(15):
... print(i, math.floor(math.log2(i+1)))
0 0
1 1
2 1
3 2
4 2
5 2
6 2
7 3
8 3
9 3
10 3
11 3
12 3
13 3
14 3
Another way to think about your nodes is via their binary representation, though in this case you should start from 1:
| i |
i+1 |
bin(i+1) |
| 0 |
1 |
0001 |
| 1 |
2 |
0010 |
| 2 |
3 |
0011 |
| 3 |
4 |
0100 |
| 4 |
5 |
0101 |
| 5 |
6 |
0110 |
| 6 |
7 |
0111 |
| 7 |
8 |
1000 |
| 8 |
9 |
1001 |
| 9 |
10 |
1010 |
| 10 |
11 |
1011 |
| 11 |
12 |
1100 |
| 12 |
13 |
1101 |
| 13 |
14 |
1110 |
| 14 |
15 |
1111 |
The depth is the position of the left-most '1' in the binary representation (That's what $\lfloor\log_2(n+1)\rfloor$ computes). Your node 9, for example, corresponding to the binary representation for 10, which i "1010". The left-most '1' is 3 positions to the left, so it's at depth 3.
Furthermore, the binary representation let lets you know how to find it: a "0" means go left and a "1" means go right. For example, the binary representation for your 13 is bin(13+1) = "1110". Start from the left-most "1". The rest of the pattern is "110". This means "right-right-left".
Go from the top of your tree (the 0), then right (on 2) then right (on 6) then left (on 13). Ta-da!
