GF$(3^2)$ can be constructed from GF$(3)$ is exactly the same way as $\mathbb C$ is constructed from $\mathbb R$ when complex numbers are first introduced to beginning students of mathematics. Just as the polynomial $x^2 + 1 \in \mathbb R[x]$ has no roots in $\mathbb R$, the polynomial $x^2 + 1 \in$ GF$(3)[x]$ has no roots in GF$(3)$. In both cases, one adjoins an element $i = \sqrt{-1}$ --- a root of the polynomial $x^2 + 1$ --- to the base field, and then creates new elements by the field operations of addition and multiplication applied to the adjoined element $i$. The properties of commutativity and associativity of addition and multiplication as well as the distributivity of multiplication over addition are assumed to hold in the field thus extended. In the case of $\mathbb R$, these operations lead to the creation of elements of the form $a + bi ~ a, b \in \mathbb R$ resulting in the complex field $\mathbb C$. Higher powers of $i$ are not necessary since $i^2 = -1 \in \mathbb R$ and thus any $i^n$ can be reduced to an element of the set $\{+1, -1, +i, -i\}$. Similarly, adjoining $i$ to GF$(3)$ gives the field GF$(3^2)$ whose nine elements are of the form $a + bi, ~ a, b \in \text{GF}(3)$. Note that $(3-a) + (3-b)i = -a -bi$ is the (unique) additive inverse of $a + bi$. Just as $a + bi \in \mathbb C$ has (unique) multiplicative inverse $(a - bi)/(a^2 + b^2) \in \mathbb C$, the (unique) multiplicative inverse of $a + bi \in \text{GF}(3^2)$ is $(a - bi)/(a^2 + b^2)$. Note that $a/(a^2 + b^2)$ and $-b/(a^2 + b^2) = 2b/(a^2 + b^2)$ are elements of GF$(3)$. In this formulation, each element of GF$(3^2)$ (or of $\mathbb C$) is described as a polynomial (of degree less than $2$) in the adjoined element $i$ which is a root of a polynomial of degree $2$.
It is also possible to consider the elements of $\mathbb C$ as polynomials of degree $1$ in an indeterminate $x$. The field operations in $\mathbb C$ then are polynomial addition and multiplication modulo $x^2 + 1$. Similarly, it is possible to consider the elements of GF$(3^2)$ as polynomials of degree $1$ in an indeterminate $x$ with field operations being polynomial addition and multiplication modulo $x^2 + 1$. In this latter case, remember that the coefficients of the polynomials are being added or multiplied modulo $3$. In either field, whether one computes $\begin{align*} (a + bx)(c + dx) &= ac + (bc + ad)x + bdx^2\\ &= (ac - bd) + (bc + ad)x + bd(x^2 + 1)\\ &\equiv (ac - bd) + (bc + ad)x \bmod (x^2 + 1)\\ \end{align*}$ or one computes $\begin{align*} (a + bi)(c + di) &= ac + (bc + ad)i + bdi^2\\ &= ac + (bc + ad)i - bd\\ &= (ac - bd) + (bc + ad)i \end{align*}$ is perhaps a matter of taste, but beginning students of finite fields are often confused when they see the field regarded sometimes as polynomials in an indeterminate and sometimes as polynomials in an element of the extension field.
Turning to the question of finding other irreducible polynomials, just as $a+bi$ and $a-bi$ are conjugates in $\mathbb C$, meaning that both are roots of the same monic polynomial $x^2 - 2ax + (a^2+b^2) \in \mathbb R[x]$, $a+bi$ and $a-bi$ are conjugates in GF$(3^2)$, meaning that both are roots of the same monic quadratic polynomial $x^2 - 2ax + (a^2+b^2) = x^2 + ax + (a^2+b^2) \in \text{GF}(3)[x].$ The choices $a=b=0$ and $a=\pm 1, b=0$ are elements of GF$(3)$ and the quadratic polynomials are reducible, while choices $a=0, b=\pm 1$, $a=1, b=\pm 1$, and $a=2, b=\pm 1$ give the three irreducible polynomials listed by Andrea Mori and also helps you easily identify which element of GF$(3^2)$ is a root of which polynomial.