Why is $x^4-3x^2+18$ irreducible over the $3$-adics? Let $f = x^4 - 3x^2 + 18 \in \mathbb{Q}_3[x]$. Since there is an LMFDB page of an extension defined by this polynomial, I assume that $f$ is irreducible. Could you verify if this is true and also why that is (not) the case?
It is not an Eisenstein polynomial (as $3^2$ divides $18$, the constant coefficient of $f$). I also tried the substitution $y = x^2$ and asked myself if $g = y^2 - 3y + 18$ is irreducible. Its reduction mod $3$ is $y^2$ which is not irreducible, so I cannot say whether $g$ is irreducible or not.

If it's reducible, it has a factor of degree $1$ or $2$.  I'll leave it up to you to
show there is no factor of degree $1$ (i.e. no root in $\mathbb Q_3$).  So suppose
$x^2 + \alpha x + \beta$ divides $x^4 - 3 x^2 + 18$.  The remainder on this division
is $$(-\alpha^3 + 2 \alpha \beta + 3 \alpha) x - \alpha^2 \beta + \beta^2 + 3 \beta + 18$$
If $\alpha = 0$ we'd have $\beta^2 + 3 \beta + 18 = 0$, which has no root in $\mathbb Q_3$,
so $\alpha \ne 0$ and $\beta = (\alpha^2-3)/2$.  Substituting in to the remainder,
this gives us $$-\frac{\alpha^4}{4} + \frac{3}{2} \alpha^2 + \frac{63}{4} = 0$$
which has no solution in $\mathbb Q_3$.
EDIT: Here's one way to see $\beta^2 + 3 \beta + 18=0$ has no solution in $\mathbb Q_3$. If $\beta$ has $3$-adic order $m$, then $\beta^2$ has order $2m$ and $3\beta$ has order $m+1$, while of course $18$ has order $2$.
So the only possibility is $m=1$.
Thus we have the $3$-adic expansion $\beta = \beta_1 3^{1} + \beta_2 3^{2} + \ldots$, with $\beta_i \in \{0,1,2\}$.  But these three possibilities make $\beta^2 + 3\beta + 18 = 
2 \cdot 3^2 + \ldots, 1 \cdot 3^2 + \ldots, 2 \cdot 3^2 + \ldots$ respectively, and not $0$.
