This answer is similar to the others; perhaps it will help to see the same
points made by yet another person.
First of all, it might help to note that $\mathbb C[x,y,z]/(xz-y)$ is
isomorphic to $\mathbb C[x,z]$. So you are looking at the map $\mathbb C[x,y]
\to \mathbb C[x,z]$ defined by $x \mapsto z, y \mapsto x z$, and asking why it is not flat.
Geometrically, this is the map $\mathbb A^2 \to \mathbb A^2$ defined by
$(x,z) \mapsto (x,xz)$. Note that a whole line in the first copy of $\mathbb A^2$ in the source (the line where $x = 0$) is mapped to a single point of the target
(the point $(0,0)$), whereas the map is an open immersion on the complement
of this line. Since open immersions are flat, this says that the point $(0,0)$
in the target is where we should focus our attention when looking for non-flatness.
(Here is a translation of my remark about open immersions in algebraic terms: if $f$ is any polynomial in $\mathbb C[x,y]$ with zero constant term, then the map on localizations $\mathbb C[x,y]_f \to
\mathbb C[x,z]_f$ is flat --- check this!)
There is one ideal that is particularly "sensitive" to the point $(0,0)$,
namely its corresponding maximal ideal $(x,y) \subset \mathbb C[x,y]$.
So let's try this ideal.
We have to look at the induced map $(x,y)\otimes \mathbb C[x,z]\to \mathbb C[x,z]$. The very equation $y = x z$ defining the map $\mathbb C[x,y]
\to \mathbb C[x,z]$ suggests an element in the kernel, namely the element
$y\otimes 1 - x\otimes z$. I leave it as an exercise to check that this element
is non-zero in $(x,y)\otimes\mathbb C[x,z]$. (If you don't know how to
make this sort of computation, then you should probably ask it as a separate
question --- but try first!)
One lesson to draw from this is that the geometry of the situation informs the algebra. A more specific remark is that the map $\mathbb A^2 \to \mathbb A^2$ from your question
is an affine patch of the blow up of $\mathbb A^2$ at the origin, and this example illustrates the general fact that non-trivial blow-ups are never flat.
Added: Looking over the other answers, it seems that one of the points of
the question is to really check that $y \otimes 1 - x \otimes z$ is non-zero
in $(x,y)\otimes \mathbb C[x,z]$.
There is a standard way to compute tensor products: by generators and relations.
While there can be other tricks in particular cases (see e.g. Michael Joyce's answer), it might be worth explaining this standard approach, since it doesn't
require any cleverness; you can always just do it.
We have to begin with a presentation of the ideal $(x,y)$ as a $\mathbb C[x,y]$-module. This is easy: it has two generators, $x$ and $y$, which
satisfy the relation $y x - x y = 0$. So we have the presentation
$$ 0 \to \mathbb C[x,y] \cdot e \to \mathbb C[x,y]\cdot f_1 \oplus \mathbb C[x,y]\cdot f_2 \to (x,y) \to 0,$$
where $e$, $f_1$, and $f_2$ are just names for basis elements of free modules,
and the maps are given by $e \mapsto (y f_1, -x f_2)$, and $f_1\mapsto x, f_2 \mapsto y$.
Now we tensor with $\mathbb C[x,z]$, to obtain the presentation
$$ \mathbb C[x,z] \cdot e \to \mathbb C[x,z] \cdot f_1 \oplus \mathbb C[x,z]
\cdot f_2 \to (x,y)\otimes \mathbb C[x,z] \to 0,$$
where again the maps are given by
$e \mapsto (y f_1, -x f_2) = (x z f_1, - x f_2)
= x(z f_1,-f_2),$ and $f_1 \mapsto x, f_2 \mapsto y = x z$.
(Note that in this particular case this exact sequence is also
exact on the left, but that is not a general feature of this approach
to computing tensor products, since generally tensoring is right-exact,
but not exact.)
From this presentation of $(x,y)\otimes \mathbb C[x,z]$, we see that
$x\otimes z - y$ (which is the image of $(z f_1, -f_2)$) is non-zero,
since $(z f_1, -f 2)$ is not in the image of the map from
$\mathbb C[x,z]\cdot e.$
On the other hand, it is a torsion element --- it is killed by multiplication by $x$ (since $x(z f_1, -f_2)$ is in the image of $\mathbb C[x,z] \cdot e$; indeed
it is the image of $e$).
This reflects the fact that if we localize away from $x = 0$ (i.e. invert
$x$), the original map becomes flat, and so the map $(x,y)\otimes\mathbb C[x,z] \to \mathbb C[x,z]$ must become injective after inverting $x$; hence its
kernel must consist of $x$-torsion elements.
Best Answer
It is because $A\simeq\operatorname{Im}f$ and tensoring an isomorphism always gives an isomorphism. Hence it is equivalent to prove that tensoring the inclusion morphism or tensoring the original injective morphism results in an injective morphism.
Btw, any rings of fractions is a flat $A$-module.