I want to show that $\mathbb{C}[s, t, u] \to \mathbb{C}[x, y] : s \mapsto x^2, t \mapsto xy, u \mapsto y^2$ is not flat.
If $s \otimes y \neq t \otimes x$ in $(x^2, xy, y^2) \otimes_{\mathbb{C}[s, t, u]} \mathbb{C}[x, y]$, then we can show it.
But I have no idea how I can show inequality of elements of a tensor.
EDIT
I'm sorry, I have misunderstood.
I want to show that the inclusion $\mathbb{C}[x^2, xy, y^2] \to \mathbb{C}[x, y]$ is not flat.
In this case, it seems that the arguments of 2 comments do not work.
So I think that I must show $x^2 \otimes y \neq xy \otimes x$ in $(x^2, xy, y^2) \otimes_{\mathbb{C}[x^2, xy, y^2]} \mathbb{C}[x, y]$.
I have seen this. But I can't find nice $\mathbb{C}[x^2, xy, y^2]$-linear map $(x^2, xy, y^2) \times \mathbb{C}[x, y] \to M$.
One of the standard way such things are done is as follows.
I state a standard theorem whose proof can be found in most text books which touches upon flatness and homological algebra.
Let $A\to B$ be a flat and finite map of Noetherian rings. Then $B$ is $A$-projective.
Assuming this, we get $B=\mathbb{C}[x,y]$ is a projective module over $A=\mathbb{C}[x^2,xy,y^2]$ if you assume flatness. Easy to check that rank of $B$ as an $A$-module is $2$ and then $B/MB$ is a $\mathbb{C}$ -vector space of dimension two for any maximal ideal $M\subset A$. But, $B/MB$ has dimension three, where $M=(x^2,xy,y^2)$, leading to a contradiction.
If you knew that a Noetherian local ring is regular if and only if its global dimension is finite, you could also argue that if $A\to B$ is finite and flat with $B$ regular, then so is $A$, giving you another argument for what you want.