Hint: Our expression is equal to $\frac{1}{x^n}(x^2+x+1)^n$.
The number of terms is the same as the number of terms in $(x^2+x+1)^n$.
Imagine expanding $(x^2+(x+1))^n$ using the Binomial Theoremm $a=x^2$, $b=x+1$. We will get $x^{2n}$ plus a bunch of lower degree terms.
How many terms? There can't be any more than $2n+1$ terms, because each term will have shape a constant times $x^k$, where $k$ is any of $0, 1, 2,\dots,2n$.
Persuade yourself, and the reader, that there are no "missing" terms, that is, no coefficient of an $x^k$, $k=0$ to $2n$, is equal to $0$. Here we are helped in the argument by the fact that the coefficients of $x^2+x+1$ are positive, so on multiplication nothing can cancel.
Added: Your approach also works well, but some detail has to be filled in. We get a large sum. Before we simplify, they are of shape $x^i\left(\frac{1}{x}\right)^j 1^k$, where $i+j+k=n$. As you pointed out, when we simplify, some of these combine. For example, $x^1(1/x)^1(1)^{n-2}$ will become $1$, as will quite a few others, when $n$ is at all large. And there will be several ways to end up with an $x^1$, also several ways to end up with a $(1/x)$, which I will call $x^{-1}$.
What is the biggest power of $x$ that we get? By choosing $i=n, j=0,k=0$ we get $x^n$. By choosing $i=0,j=n,k=0$ we get $x^{-n}$. That's the smallest (most negative) power that we get. Between $n$ and $-n$, inclusive, we have $2n+1$ numbers (don't forget $0$).
We want to see that we get a term in every one of these powers. By symmetry, it is enough to show that we get stuff in every non-negative power of $x$, from the $n$-th power to the $0$-th power.
For $0\le m\le n$, we get $x^m$ in the expansion, for example by using $i=m,j=0,k=0$, and usually in several other ways. But the point is we do get some stuff that involves $x^m$. So every power of $x$, from $x^n$ down to $x^{-n}$ is represented, a total of $2n+1$.
Since the size of the problem is small, we can count the cases directly. First, there are two partitions of 5 into at most 4 parts with each part at most 2, namely, $5=2+2+1$ and $5=2+1+1+1$. These correspond to the ways you can get $x^5$.
The partition $5=2+2+1$ means you get 2 factors of $-x^2$ from two of the terms, and a factor of $2x$ from another one. You can do this in $\binom42\binom21=12$. So you $12$ terms of the form $(-x^2)^2 2x$; in total $24x^5$.
For $5=2+1+1+1$ there are $\binom41\binom33=4$ cases and the terms are of the form $-x^2(2x) (2x)(2x)=-8x^5$; in total $-32x^5$.
So the $x^5$ term is
$$
24x^5-32x^5= -8x^5.
$$
In your method you also need to expand $(2x-x^2)$ with a binomial sum.
Note: Corrections made based on the comment below.
Best Answer
Hint: The binomial sum is $$\sum_{k = 0}^{100} {100 \choose k} 3^{k/5}7^{(100-k)/3},$$ and the rational terms are exactly the integer terms, i.e., when both $k/5$ and $(100-k)/3$ are integers.