Let $R=k[x_0,...,x_n]$ and $R_j=k[\frac{x_0}{x_j},...,\frac{x_{j-1}}{x_j},\frac{x_{j+1}}{x_j},...,\frac{x_n}{x_j}]$
If $f\in R$, and $f_d,...,f_m$ its homogeneous components and $d,...,m$ their degrees
We can define a map from $R$ to $R_j$ by sending $f$ to $\frac{f_d}{{x_j}^d}+...+\frac{f_m}{{x_j}^m} $
Now take a prime ideal $P$ of $R_j$, we know that $P$ must be finitely generated by $H_1,...,H_r$ polynomials in $R_J$ with degrees of $e_1,...,e_r$ respectively. Then ${x_j}^{e_i}H_i = h_i$ are homogeneous polynomials in $R$
My question is if $φ^{-1}(P)=(h_1,...,h_r)$
The one direction is obvious as $φ(h_i)=H_i$ but I'm having trouble showing the reverse direction.
The answer to your question is "no". Let me illustrate by a trivial example. Choose $n=0$, so you have only one indeterminate in $R$, say $x$, and your map reduces to $$\varphi: k[x] \rightarrow k \qquad f\mapsto f(1)$$
and the field $k$ has only one (strict) prime ideal $\{0\}$. But $\varphi^{-1}(0)$ is clearly not just $\{0\}$, because $\varphi(x-1)=0$.