Proving that for structure $M=(\mathbb R; +,<,0)$, the function $f(x,y)=x \cdot y$ is not a definable set in $M$.
Note: A function is called a definable set, if its graph, meaning the set $(x,f(x)), x \in M $ (the structure's universe) is a definable set in the structure.
So in this case we need to prove the set $(x,y,x \cdot y)$ is not definable in $M$. Using the last section in the Wikipedia article on a definable set, we know that if $X$ is definable on structure $M$ over parameters $c_1, \cdots, c_s$ then for every automorphism $\alpha: M \rightarrow M$ so that for every parameter $\alpha(c_i)=c_i$ , if $(x_1, x_2,..) \in X$ then $(\alpha(x_1), \alpha (x_2),..) \in X$ as well.
So to show that the function is not definable without parameters, we can assume it is definable, then look at the homomorphism $\alpha(x)=2x$ and see it isn't invariant, as $(2,3,6)\in X$, however $(4,6,12) \not\in X$.
What I try to do now for a really long time is proving it is not definable with parameters as well. Any help will be great!
Idea I thought of, not sure if correct. Lets assume $X$ is definable over parameters $c_1,...,c_s$. $\mathbb R$ is a vector space over $\mathbb Q$. Lets set $V_1=\text{span} \{c_1, c_2, ..., c_s\}$. So we know there is some $V_2$ so that $\mathbb R=V_1 \oplus V_2$. And for every $r \in \mathbb R$, there are unique $v_1 \in V_1, v_2 \in V_2$ so that $r=v_1+v_2$.
So lets set a homomorphism like so: $\alpha(r)=\alpha(v_1+v_2)=v_1+2v_2$. It indeed preserves $<$ and $+$, so it is a homomorphism. And for every $c_i \in V_1$, $\alpha(c_i)=c_i$, so it's invariant over the parameters. The only question is whether, for $a_1,a_2 \in V_2$, $a_1 \cdot a_2 \in V_2$. Because, if so, we can look at $\alpha(a_1 \cdot a_2)=2a_1a_2$, while $\alpha(a_1)\cdot \alpha(a_2)=4a_1a_2$, thus showing $X$ is not definable.
As hot_queen notes in the comments, you can show this using quantifier elimination. But the idea the you suggested will actually work! The problem was that any collection of parameters from $\mathbb{R}$ control all of $\mathbb{R}$ (as in David Ullrich's answer) - you don't have the flexibility of automorphisms that you'd like. The solution is to leave $\mathbb{R}$ behind and move to an elementary extension which is more flexible.
Pick some elementary extension $\langle \mathbb{R}^*;+,<,0\rangle$ of $\mathbb{R}$ which contains "infinite elements", i.e. realize the type $\{x > r\mid r\in \mathbb{R}\}$. Now let $V_1$ be the convex hull of $\mathbb{R}$ in $\mathbb{R}^*$: $$V_1 = \{x\mid \exists r_1, r_2\in\mathbb{R}, r_1 < x < r_2\}.$$ Notice that $V_1$ is not all of $\mathbb{R}^*$ (since there are infinite elements), but $V_1$ is a $\mathbb{Q}$-subspace of $\mathbb{R}^*$. So we can write $\mathbb{R}^* = V_1\oplus V_2$ and decompose any element $r\in\mathbb{R}^*$ as $r = v_1 + v_2$ with $v_1\in V_1$ and $v_2\in V_2$.
Consider, just as you suggested, the map $\alpha(r) = v_1 + 2v_2$. $\alpha$ is clearly an invertible linear map which fixed $\mathbb{R}$ but is not the identity on $\mathbb{R}^*$. And this time it respects $<$.
[Why? If $r < r'$, then writing $r = v_1 + v_2$ and $r' = v_1' + v_2'$, we have that $r' - r = (v_2' - v_2) + (v_1' - v_1) > 0$. Since $v_2' - v_2\notin V_1$, it's either positive and greater than everything in $V_1$ or negative and less than everything in $V_1$. Well, it must be positive since the sum is positive, so also $\alpha(r') - \alpha(r) = 2(v_2' - v_2) + (v_1'-v_1) > 0$, and $\alpha(r) < \alpha(r')$.]
Now suppose multiplication is definable in $\mathbb{R}$. We can use the same formula to define a multiplication function on $\mathbb{R}^*$ with the same first-order properties (since $\mathbb{R}^*$ is an elementary extension). Take $r\in V_2$ with $r>0$. Since multiplication by a positive element is an increasing function, and $r$ is greater than every element of $\mathbb{R}$, the same is true of $r^2$. Then $(\alpha(r))^2 = (2r)^2 = 4r^2$, but $\alpha(r^2) \leq 2r^2$.