Is it ok this CNF of a Boolean function? I have to find out the CNF of $$\begin{matrix} f(x,y,z)&=&(x\wedge y)\vee(x\wedge z),\end{matrix}$$ where $f$ is a Boolean function.

$$\begin{matrix}&f(x,y,z)&=&(x\wedge y)\vee(x\wedge z)&1\\
&&=&x\wedge(y\vee z)&2\\
&&=&(x\vee0_B\vee0_B)\wedge(y\vee z\vee 0_B)&3\\
&&=&(x\vee(y\wedge\overline y)\vee(z\wedge\overline z))\wedge(y\vee z\vee(x\wedge\overline x))&4\\
&&=&(((x\vee y)\wedge(x\vee\overline y))\vee(z\wedge\overline z))\wedge((y\vee z\vee x)\wedge(y\vee z\vee\overline x))&5\\
&&=&(x\vee y\vee z)\wedge(x\vee\overline y\vee\overline z)\wedge(x\lor y\lor z)\wedge(\overline x \lor y\lor z)&6\\
&&=&(x\vee y\vee z)\wedge(\overline x\vee y\vee z)\wedge(x\vee\overline y\vee\overline z),\end{matrix}$$
where $0_B$ is the first element.
I have to make sure that all the terms refer to all variables involved.
Anyway WolframAlpha says another thing... What am I doing wrong?

In general, formulas do not have a unique conjunctive normal form (CNF), see an example here. So, in general, the fact that you get another CNF does not imply automatically that you are wrong. The important thing is that all CNF's you get should be equivalent.
In this particular case, you should notice that actually in line 2 (after applying the distributivity law to line 1) you already have a CNF $x \land (y \lor z)$ and so you can stop there.
Moreover, the formula in your last line $(x\vee y\vee z)\wedge(\overline x\vee y\vee z)\wedge(x\vee\overline y\vee\overline z)\,$ is not equivalent to $x \land (y \lor z)$ (indeed, consider the truth assignment $v$ where $v(x) = v(y) =\bot$ and $v(z) = \top$). This means that you actually did something wrong and $(x\vee y\vee z)\wedge(\overline x\vee y\vee z)\wedge(x\vee\overline y\vee\overline z)\,$ is not a CNF of $(x\wedge y)\vee(x\wedge z)$.
Your mistake is between lines 5 and 6. You should write:
\begin{align}
((x \lor y) \land (x \lor \overline{y})) \lor (z \land \overline{z}) \dots  &= ((x \lor y) \lor (z \land \overline{z})) \land ((x \lor \overline{y}) \lor (z \land \overline{z})) \dots  \\
&= (x \lor y \lor z) \land (x \lor y \lor \overline{z}) \land (x \lor \overline{y} \lor z) \land (x \lor \overline{y} \lor \overline{z}) \dots
\end{align}
Therefore, a CNF of $(x\wedge y)\vee(x\wedge z)$ where all clauses contain exactly one occurrence of each variable is 
\begin{equation}\tag{1}
 (x \lor y \lor z) \land (x \lor y \lor \overline{z}) \land (x \lor \overline{y} \lor z) \land (x \lor \overline{y} \lor \overline{z}) \land (\overline{x} \lor y \lor z)
\end{equation}
Using truth tables, you can easily prove that formula $(1)$ is equivalent to $x \land (y \lor z)$, so both are CNF of $(x\wedge y)\vee(x\wedge z)$.
