Let $f:X\to Y$ be the normalization of a curve $Y$ with a node. Then this example 9.7.1 (Hartshorne's book Algebraic Geometry) want to show $f$ cannot be flat. But I don't understand the following step:
If $f$ were flat then $f_* \mathcal O_X$ would be a flat sheaf of $\mathcal O_Y$ modules.
Then the Definition in page 254 tells that this means, for each point $x\in X$, the local ring $(f_*\mathcal O_{X})_x$ is a flat $O_{Y,f(x)}$-module. (Actually, the definition of flat sheaves is not pointed out in the book, but we may define it in analogy to the one in page 254) On the other hand, the same definition tells that the flatness of $f$ can only implies that $\mathcal O_{X,x}$ is a flat $\mathcal O_{Y,f(x)}$-module via $f^\#: \mathcal O_{Y,f(x)}\to \mathcal O_{X,x}$, slightly different from the previous one.
In general, two stalks you mentioned are different.
Here you have an additional assumption: $f$ is an affine morphism. Let $V=\mbox{Spec}~A$ be an affine open set of $Y$ and $f^{-1}(V)=\mbox{Spec}~B$, where $B$ is the integral closure of $A$. If $\mathfrak{p}$ represents $x$ and $\mathfrak{p}\cap A=\mathfrak{q}$ represents $y$, then
$\mathscr{O}_{X,x}\cong B_\mathfrak{p}$, $(f_*\mathscr{O}_X)_{f(x)}\cong B\otimes A_\mathfrak{q}$, and $\mathscr{O}_{Y,y}\cong A_\mathfrak{q}$.
You are asking whether or not $B\otimes A_\mathfrak{q}$ is flat over $A_\mathfrak{q}$.
Suppose $f$ is flat, i.e., $B_\mathfrak{p'}$ is flat over $A_\mathfrak{q'}$ (Here $\mathfrak{q'}=A\cap\mathfrak{p'}$) for any $\mathfrak{p'}\in\mbox{Spec}~B$. This is equivalent to that $B$ is flat over $A$. Hence $B\otimes A_\mathfrak{q}$ is flat over $A_\mathfrak{q}$.