$\mathbb{C}^2$ is a group over field $\mathbb{C}$, with the following actions: addition is similar to the regular addition. multiplication is defined by:
for every $(z,w) \in \mathbb{C}^2$ and every $\alpha \in \mathbb{C}$. $$\alpha \cdot (z,w) = (\alpha \overline{z}, \overline{\alpha}w)$$
I found out that addition rules apply here, but the associativity of the multiplication was not applying. that's how I found it:
$$w,z,\alpha,\beta \in \mathbb{C}, (w,z) \in \mathbb{C}^2 $$ From the first side: $$(\alpha\beta)(w,z) = (\alpha\beta\overline{z}, \overline{\alpha\beta}w)$$ But on the other hand: $$(\alpha\beta)(w,z)=\alpha(\beta(w,z)) = \alpha(\beta\overline{z}, \overline{a}w) = (\alpha\overline{\beta}z,\alpha\overline{\beta}w)$$ so associativty is not applying here.
But I'm really not sure about my answer. is that how you prove it?
(Comment turned answer, and expanded a bit)
Your proof is good, though I would refrain from writing $(\alpha\beta)(z,w)=\cdots$ in the last line because, as you've shown, it isn't true. Alternatively, you could say something to the effect of "Suppose for the sake of a contradiction that $(\alpha\beta)(z,w)=\cdots$" and then derive the contradiction.
Strictly speaking, you should actually give an actual example to show that $(\alpha\overline{\beta}z,\alpha\overline{\beta}w)=(\alpha\beta\overline{z},\overline{\alpha\beta}w)$ does not hold, but such examples are not difficult to find (take, for example, $z=w=1$ and $\alpha=\beta=i$, so that you get $(1,1)$ on the LHS and $(-1,-1)$ on the RHS).
An easier way to show that $\mathbb{C}^2$ does not form a $\mathbb{C}$-vector space under the given scalar multiplication is by noticing that compatibility with $1$ does not hold: $1\cdot (z,w)=(\overline{z},w)$, so by choosing $z$ to be non-real, we see that $1\cdot (z,w)=(z,w)$ does not hold in general.