
sorry for uploading weird angle pictures..but no other concise ways I can't think of..
this is the p.49 of Klaus Hulek elementary algebraic geoemtry on the last paragraph it says that K[V] is not a UFD, so the representation of f is not unique. What does it mean? does it mean that even if g/h = d/f, gf=dh does not hold?
It means just what it says: the same element of $k(V)$ can be represented as a fraction in different ways. (The question in your last sentence has a negative answer, because that is the definition of equality in $k(V)$.)
Let's spell it out a bit more. Suppose that an element $f \in k(V)$ can be written as a fraction in two different ways: $f=\frac{g_1}{h_1}=\frac{g_2}{h_2}$. (Let's suppose that $g_1$ and $h_1$ have no common irreducible factor, and the same for $g_2$ and $h_2$.) By definition of equality in $k(V)$, this means that
$$g_1 h_2 = g_2 h_1.$$
Now if $k[V]$ is a UFD, the only way this can hold is if $g_1$ and $g_2$ are associates, and the same for the $h_i$. So in this case, up to units, our representation of $f$ is unique.
On the other hand, if $k[V]$ is not a UFD, this is no longer true. For example let $k[V]=k[x,y,z]/I$ where $I$ is the ideal generated by $z^2-xy$. Then in $k[V]$ we have the equality $$z^2=xy$$ but $z$ is not an associate of either $x$ or $y$. Turning this into a statement about rational functions, we get the equality of rational functions $$\frac{z}{x}=\frac{y}{z}$$ where the numerators on either side, and the denominators, are genuinely different. So here is a rational function on $V$ with a non-unique representation.