The graph factorizes as
$$
p(x)=p(x_2)p(x_1\mid x_2)p(x_4\mid x_2)
p(x_3\mid x_1,x_4)p(x_5\mid x_2)
p(x_6\mid x_4,x_5).
$$
Conditioning on all variables except $X_1$, terms not involving $x_1$ cancel, leaving
$$
p(x_1\mid x_2,x_3,x_4,x_5,x_6)
\propto p(x_1\mid x_2)p(x_3\mid x_1,x_4).
$$
This depends only on $(x_2,x_3,x_4)$, so
$$
X_1\perp(X_5,X_6)\mid(X_2,X_3,X_4).
$$
Thus $(X_2,X_3,X_4)$ is a <Markov blanket> of $X_1$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 221 / 3 / i / Solution

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A Markov blanket of a variable is a set that makes it conditionally independent of every remaining variable. In a <Directed acyclic graph>, the parents, children, and other parents of the children form the standard Markov blanket.

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 Incoming links: Markov blanket