Let $_9$ be constructed with the irreducible polynomial $x^2 +2x+2$.
For $a,b \in _3$ we write $ax+b \in _9$ for $ab$.
In our exam we had to find out whether the following are true or false. I know the answers, but I don't understand how one can find out.
$$22 \text{ is the inverse of } 11 \text{ -> True}$$ $$22 \text{ is the negative of } 11 \text{ -> True}$$
$$\text{For random } a \in _9 \setminus \{00\} \text{ it holds that } a^3 =01 \text{ -> False}$$ $$\text{For random } a \in _9 \setminus \{00\} \text{ it holds that } 3 \cdot a =01 \text{ -> False} $$
I just know that $_3[x](x^2+1)$ is a finite field with $9$ elements and $_9$ is isomorphic to it, but what is the approach when trying to figure out which of the above are true/false?
If the polynomial was $x^2+x+2$, then the answers would be False, True, False, False.
Note that $$\Bbb{F}_9=\Bbb{Z}_3[x]/\langle x^2+2x+2 \rangle=\{ax+b \, | \, a,b \in \Bbb{Z}_3 \text{ and } x^2+2x+2 \equiv 0\}.$$ The notation you are using is a tad confusing: when you write $22$, I am assuming it refers to $2x+2$.
To check if $22=2x+2$ and $11=x+1$ are inverses of each other, we do the following: \begin{align*} (2x+2)(x+1)&=2(x+1)^2\\ &=2(x^2+2x+1)\\ &=2(-2+1) && (\because x^2+2x+2 \equiv 0)\\ &=-2\\ &=1 && (\because -2 \equiv 1 \text { in } \Bbb{Z}_3) \end{align*} Thus $2x+2$ and $x+1$ are inverses.
Added information:
Let $\alpha=ax+b \in \Bbb{F}_9 \setminus \{0\}$. Suppose $\alpha^3 =1$ for all such $\alpha$'s, then that would mean the order of all non-zero and non-identity elements is $3$. But the order of the multiplicative group $\Bbb{F}_9^{\times}=\Bbb{F}_9\setminus \{0\}$ is $8$ and $3$ does not divide $8$, so this is not possible by Lagrange's Theorem.
But here is an alternate way (simple calculation) to check this without invoking Lagrange's theorem
Let's take $\alpha=x$ (in your notation it is $10$). Then \begin{align*} x^3 & = x(x^2)\\ &=x(-2x-2) && (\because x^2+2x+2 \equiv 0)\\ &=x^2+x\\ &=-2x-2+x\\ &=-x-2\\ & =2x+1. \end{align*} THus $x^3 \not \equiv 1$.
Hopefully you can take the rest from here.