I think the question is rather about what is going on, than about "solving" the problem. So let us compare the given problem with the following one, invented for the reason of being similar with a simpler arithmetic situation:
Compute $2/4\in\Bbb Z/(7)$.
There is no profit in knowing here the answer for the stated question, but the structure is similar. So i will explain this structure first. On the R.H.S. we have a ring modulo an ideal, the ring is the ring of integers, and the ideal is the ideal generated by $7$, so we consider to be zero (in the quotient ring) elements that are multiples of $7$, we simply write $0=\pm7=\pm 14=\pm21=\dots$ in the ring $R=\Bbb Z/(7)$. (Which is by chance a field.) Now we have to make a computation in this ring, what is $2/4$ in $R$? Well, we can claim it is simply $2/4$, which is an element in $R$, or simplify and claim it is $1/2$. Both answers are valid. But implicitly, tacitly the question wants from us to have an answer using the "standard representatives" from $R=\Bbb Z/(7)$, which are (depending on taste) either the seven (classes of the) elements $0,1,2,3,4,5,6$, or the other used set $0,\pm1,\pm2,\pm3$. Well, we have a finite set of candidates, and the answer is quickly find by trial and error, it is $4$, why? - well we check that $2/4=4$ in $\Bbb R$, equivalently $2=4\cdot 4$ in $R$, equivalently $2=16$ in R$, equivalently $2-16$ is a multiple of seven, and this is indeed the case.
We translate in the given situation. The ring is (no longer $\Bbb Z$), but the ring $\Bbb F_7[X]$ of polynomials in the indeterminate $X$ with coefficients in the field with seven elements $\Bbb F_7$. The ideal we consider in this ring is the ideal generated by the one polynomial $X^3+X^2+1$. We are searching for a better representative for $2X^2/(2X^2+2)$ in the quotient ring. Again, the problem wants (tacitly) us to find a "good representative". (Modulo seven, the possible rests when dividing with rest by seven are $0,1,2,3,4,5,6$. Which are the possible rests when dividing with rest by a polynomial of third degree?) We search thus the answer in the form $aX^2+bX+c$ with coefficients $a,b,c\in\Bbb F_7$. The check is this time:
$$
\frac {2X^2}{2X^2+2}=aX^2+bX+c\in\Bbb F_7[X] \text{ modulo }(X^3+X^2+1)
\\
\text{ iff }
\\
2X^2-(2X^2+2)(aX^2+bX+c)\text{ is a multiple of }(X^3+X^2+1)
\\
\text{ iff }
\\
(X^2+1)(aX^2+bX+c)\text{ is equal to $X^2$ modulo }(X^3+X^2+1)
\ .
$$
We compute thus $(X^2+1)(aX^2+bX+c)$ modulo $(X^3+X^2+1)$ as a polynomial of degree $\le 2$ in $X$, and finally insist this polynomial is $X^2$.
$$
\begin{aligned}
(X^2+1)(aX^2+bX+c)
&=
aX^4+bX^3+(c+a)X^2+bX+c\\
&\equiv
(aX^4+bX^3+(c+a)X^2+bX+c)-aX(X^3+X^2+1)\\
&=
(b-a)X^3+(c+a)X^2+(b-a)X+c\\
&\equiv
(b-a)X^3+(c+a)X^2+(b-a)X+c-(b-a)(X^3+X^2+1)\\
&=(c-b+2a)X^2+(b-a)X+(c-b+a)\ .
\end{aligned}
$$
The last expression is set to be equal to $X^2$, so we solve the system with three equations $c-b+2a=1$, and $b-a=0$, and $c-b+a=0$, to obtain $a=b$, then $c=0$, then $a=1$, so the answer is $X^2+X$.
Note: If we know this answer, then the whole story reduces to the check:
$2X^2/(2X^2+2)$ is $X^2+X$ because $(2X^2+2)(X^2+X)=2X^4+2X^3+2X^2+2X=2X\color{blue}{(X^3+X^2+1)}+2X^2\equiv 2X^2$
modulo $\color{blue}{(X^3+X^2+1)}$.
Note: Computers may be useful to understand similar computations and check results. For instance, the result is obtained as follows in sage:
sage: R.<X> = PolynomialRing(GF(7))
sage: R
Univariate Polynomial Ring in X over Finite Field of size 7
sage: Q.<x> = R.quotient(X^3 + X^2 + 1)
sage: 2*x^2/(2*x^2 + 2)
x^2 + x
The computation in the other quotient ring:
sage: R.<X> = PolynomialRing(GF(7))
sage: R
Univariate Polynomial Ring in X over Finite Field of size 7
sage: Q.<x> = R.quotient(X^3 + X + 1) # the other quotient ring
sage: Q.is_field()
True
sage: 2*x^2/(2*x^2 + 2)
x + 1
Again: Knowing the answer makes the solution a simple check:
$$(2X^2+2)(X+1)=2X^3+2X^2+2X+2=2\color{blue}{(X^3+X+1)}+2X^2\equiv 2X^2$$
modulo $\color{blue}{(X^3+X+1)}$.
