Let $\theta$ be a root of $p(x)=x^3+9x+6$, find the inverse of $1+\theta$ in $\mathbb{Q(\theta)}$

Correct: it takes only $1$ step (= division) in the extended Euclidean algorithm to invert a linear polynomial $\,g\,$ since $ f = q\,g + c \,\Rightarrow\, \bmod f\!:\ q\,g\equiv -c\,\Rightarrow\, 1/g \equiv -q/c,\,$ just as you calculated.

---

Generally it's easier to use said augmented-matrix form of the extended Euclidean algorithm, e.g. below we compute $\,1/g \pmod{\!f} = 1/(x^2\!+\!1) \pmod{\!x^3\!+\!2x\!+\!1}\,$ over $\,\Bbb Z_3,\,$ from this answer.

$\begin{eqnarray} [\![1]\!]&& &&f = x^3\!+2x+1 &\!\!=&\, \left<\,\color{#c00}1,\,\color{#0a0}0\,\right>\quad\ \ \, {\rm i.e.}\ \qquad\!\:\! f\, =\ \color{#c00}1\cdot f\, +\, \color{#0a0}0\cdot g\\ [\![2]\!]&& &&\qquad\ \, g =x^2\!+1 &\!\!=&\, \left<\,\color{#c00}0,\,\color{#0a0}1\,\right>\quad\ \ \,{\rm i.e.}\ \qquad g\, =\ \color{#c00}0\cdot f\, +\, \color{#0a0}1\cdot g\\ [\![3]\!]&=&[\![1]\!]-x[\![2]\!]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! &&\qquad\qquad\ \ x+1 \,&\!\!=&\, \left<\,\color{#c00}1,\,\color{#0a0}{-x}\,\right>\ \ \ \:\!{\rm i.e.}\quad\! x\!+\!1\, =\, \color{#c00}1\cdot f\,\color{#0c0}{-\,x}\cdot g\\ [\![4]\!]&=&[\![2]\!]+(1\!-\!x)[\![3]\!]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! &&\qquad\qquad\qquad\ 2 \,&\!\!=&\, \left<\,\color{#c00}{1\!-\!x},\,\ \color{#0a0}{1\!-\!x+x^2}\,\right>\\ \end{eqnarray}$

Hence the prior line implies: $\ \ 2\ =\ (\color{#c00}{1\!-\!x})f + (\color{#0a0}{1\!-\!x\!+\!x^2})g $

Thus in $\,\Bbb Z_3[x] \bmod f\!:\,\ {-}1\equiv 2 \equiv (\color{#0a0}{1\!-\!x\!+\!x^2})g\ \Rightarrow\ \bbox[6px,border:1px solid red]{g^{-1}\equiv\, {-}(\color{#0a0}{1\!-\!x\!+\!x^2})}$