How does one show $(\lambda x . (\lambda y.x))yx \equiv_{\beta} y$ in lambda calculus?
I will make my answer more sophisticated by using parentheses. The first thing to realize is that $(\lambda x . (\lambda y . x)) y x$ is the same thing as $((\lambda x . (\lambda y . x)) y) x$ because in $\lambda$-calculus application associates to the left, which means that $A B C = (A B) C$. As you can see, $x y$ does not even appear in your expression.

After that, I would follow Yuval's observation that it is impolite to use the same variable for two different purposes: the $x$ and $y$ inside $\lambda$'s are not the same as the outer $x$ and $y$. We should rename bound variables to get
$$((\lambda a . (\lambda b . a)) y) x$$
Can you do it from here?

P.S. Please stop pelting us with questions about $\lambda$-calculus that all look alike. Take them one at a time, wait for the answers, and think about them. It is likely subsequent questions will become redundant.