Find inverse element of $1+2\alpha$ in $\mathbb{F}_9$ 
Let $$\mathbb{F}_9 = \frac{\mathbb{F}_3[x]}{(x^2+1)}$$ and consider $\alpha = \bar{x}$. Compute $(1+2 \alpha)^{-1}$

I think I should use the extended Euclidean algorithm: so I divide $x^2 +1 $ by  $(1+2x)$:
$$x^2 + 1 = (1+2x)(2x+2)+2$$
$$(2x+2)(1+2x) + 2(x^2+1) = 1$$
Therefore, considering $\text{mod}(x^2+1) $, I have $$(2x+2)(1+2x) = 1\text{mod}(x^2+1)$$
and so $2x+2 = (1+2x)^{-1}$
Is it okay, or did I misunderstood something?

Your answer is correct.
Note that $\alpha^2=-1$, and the conjugate of $\alpha$ is $-\alpha$,
so $\dfrac1{1+2\alpha}=\dfrac1{1+2\alpha}\dfrac{1-2\alpha}{1-2\alpha}=\dfrac{1-2\alpha}{1-4\alpha^2}=\dfrac{1-2\alpha}{1-\alpha^2}=\dfrac{1-2\alpha}2=2^{-1}-\alpha=2+2\alpha$
in $\mathbb F_9$.
