This is the a-part of problem 11 from the book Ideals, Varieties and Algorithms (4th ed., pp. 142).
Let $k$ be a field and define the rational parametrization $x_i = \frac{f_i(t_1,\dots,t_m)}{g_i(t_1,\dots,t_m)}$ for $1 \leq i \leq n$ where $f_i, g_i \in k[x_1,\dots,x_n]$ are polynomials for $1 \leq i \leq n$. Define $g = \prod_{i=1}^ng_i$. I have to show that for any $h \in k[x_1,\dots,x_n]$ there exists sufficiently large $N$ and $F \in k[t_1,\dots,t_m,x_1,\dots,x_n]$ such that $g^Nh = F(t_1,\dots,t_m,g_1x_1,\dots,g_nx_n)$.
Currently, I have no idea how to begin, because I don't understand the problem well enough: If we take $h = x_1 + 1$, then according to the statement, we could find a large enough $N$ and some function $F$ such that $g^N(x_1 + 1) = F(t_1,\dots,t_m,g_1x_1,\dots,g_nx_n) \Longleftrightarrow x_1 + 1 = g^{-N}F(t_1,\dots,t_m,g_1x_1,\dots,g_nx_n)$. We see immediately that if $N = 0$, then we are in trouble as any non-constant polynomial $F$ will contain some product of $g_ix_i$s. Thus we would need $N \neq 0$ in order to "clear" the $g_1$ factor from $g_1x_1$. But if, say, $N = 1$, then $F$ would need to have a factor $\prod_{i=1}^ng_ix_i$ in order to cancel out the denominator $\prod_{i=1}^ng_i$. But then $F$ would have a factor of $x_1x_2\cdots x_n$, and I don't really see how you can make any of $x_j, 2 \leq j \leq n$ disappear.
So how should I approach this problem?
We may assume that $h$ is a monomial. Let's consider a particular case: $h(x_1,x_2)=x_1^2x_2^3$. Since $$x_i=f_i(t_1,...,t_m)/g_i(t_1,...,t_m)$$ we get $h=f_1^2f_2^3/g_1^2g_2^3$, and multiplying by $g^3$ we get $g^3h=f_1^2f_2^3g_1=(x_1g_1)^2(x_2g_2)^2g_1$ which is a polynomial in $x_1g_1$, $x_2g_2$, and $t_1,...,t_m$ (this is $g_1$).
If the above reasoning is clear and correct, then you know what to do in general.