I assume the $Y$ in your post is the number of rolls to obtain the first 6 and that $X$ is the number of rolls to obtain the first 1. For your problem, to find $p_{X|Y}$, you need to calculate probabilities of the the form $P[X=i|Y=3]$. You could use the formula stated in your post: find the probability that [the first 6 occurred on the third roll and the first 1 occurred on the $i$th roll], then divide by the probability that the first 6 occurred on the third roll.
But here is a simpler way of solving your problem:
Given that the first 6 occurred on the third roll:
let $X$ be the number of rolls to the first 1.
Let $A$ be the event that the first roll was 1.
Let $B$ be the event that the first roll was not 1 and the second roll was 1.
Let $C$ be the event that the first 1 occurred after the third roll.
You can use $ \Bbb E(X)= \Bbb E(X|A)P(A)+ \Bbb E(X|B)P(B) +\Bbb E(X|C)P(C) $
We have:
$P(A)={1\over5}$ and $\Bbb E(X|A)=1$.
$P(B)={4\over5}\cdot{1\over5}$ and $\Bbb E(X|B)=2$.
As for the event $C$:
$P(C)={4\over5}\cdot{4\over5}$.
To find $\Bbb E(X|C)$, note:
If the first two rolls were not 1 (and remember, we are given that first 6 occurred on the third roll), what is the expected number of rolls to obtain the first 1? Well, it would be 3 plus the expected number of additional rolls (past the third) to get a 1. But, in this situation, the expected number of additional rolls to get a 1 is the same as the expected number of rolls to obtain a 1 with no conditions, which is 6. So, if the first two rolls were not 1, the expected number of rolls to obtain the first 1 would be 3+6=9.
So, the expected number of rolls to the first 1, given that the the first 6 occurred on the third roll, is: $ 1\cdot{1\over5}+2\cdot{4\over5}\cdot{1\over5} + 9\cdot{4\over5}\cdot{4\over5}. $