Please explain this ideal quotient in $k[x,y,z,o]$:
$$\langle x\rangle : \langle x y z \rangle=\{f\in k[x,y,z,o] : fg\in \langle x \rangle\quad\forall g\in \langle x y z \rangle \}$$
where
$$fg\in\langle x\rangle \quad\forall g\in\langle xyz\rangle \leftrightarrow fg=Ax\quad \forall q\in \langle xyz\rangle$$
$$f xyz=Bx \leftrightarrow f y z= B \quad\text{(Wrong as observed by Chen)}$$
and my goal is compute manually to the result $\langle x\rangle : \langle x y z \rangle=\langle 1\rangle$.
Correct result checked with M2
"R=QQ[x,y,z,o]; ideal(x):ideal(xyz)" results to "ideal 1" so the whole polynomial ring k[x,y,z,o].
Your answer is not right.
The ideal quotient $I \colon J$ is the set of all $r \in R$ such that $rJ \subseteq I$, for $I,J$ ideals of a commutative ring $R$.
Observe that $\langle xyz \rangle \subset \langle x \rangle$ already, so for any $r \in k[x,y,z,o]$, we have that $r\langle xyz \rangle \subseteq \langle xyz \rangle \subseteq \langle x \rangle$ and hence $\langle x \rangle \colon \langle xyz \rangle$ is all of $k[x,y,z,o]$.