I've been given a two part question to first find the number of elements and then number of units is in $R/I$ where $R = \mathbb{F}_7[x]$, $f(x) = x^3 + 4$ and $I = f(x)$
I found the number of elements through my own notes, where we have $p^k$ where $p$ is the size of the ring, in this case $7$ and $k$ is the degree of the polynomial element, in this case $3$. Therefore the number of elements is $7^3 = 343$.
Now i know that a unit is an element $u \in R$ such that for any element $v \in R$, $uv = 1$, therefore a unit is an element in $R$ with an inverse also in $R$. However I'm unsure how to use this information to help find how many units there are out of the 343 elements as it would be impractical to start testing each element one by one.
In general you could do something like this:
Since $F_7[x]$ is a principal ideal domain, an element $u$ of $F_7[x]$ is going to have an inverse in $F_7[x]/(f(x))$ precisely when $lcd(u, f(x))=1$. Otherwise they would both lie in the proper ideal generated by their least common divisor.
If you can count the number of polynomials of degree less than that of $f$ that are coprime with $f$, you have the number of units.
So, a good first step would be to factor $f(x)$ over $F_7$ completely. In your case it's not too hard to verify all the residues of $x^3$ are in $\{-1,0,1\}$ mod $7$, so $x^3+4$ has no roots, and is irreducible. So it turns out all elements are coprime to $x^3+4$, and are invertible.
Note: apologies for the last version of the solution. I had muddled $x^3+4$ for $x^4+3$ at the end from a problem I had previously seen. I had thought the problem would be more challenging than just a field.
Since I started out analyzing $f(x)=x^4+3=(x-2)(x+2)(x^2+2)$ anyways, I may as well finish. It is a diverting combinatorial task to count the units. One can count the monic polynomials that aren't coprime with $f(x)$, and then multiply by $6$ (to recover the nonmonic ones.) Just do one degree at a time.
Firstly, there are (obviously) two monic polynomials of degree $1$ which have a common factor with $f(x)$.
For degree $2$, there are $14$ such polynomials: $x^2+2$ itself, the product $(x+2)(x-2)$, and $12$ more products of one of the two with some other $x-a$.
For degree $3$: for $x-2$, there are $49$ monic degree $3$ polynomials which it divides. One of these it shares with $x^2+2$, and $7$ are shared with $x+2$. Symmetrically the same thing can be said for $x+2$. Finally, $x^2+2$ can be paired with seven linear factors, and one case overlaps each of the linear factors $x+2$ and $x-2$. So if you draw up a little Venn diagram of the situation, you can see its total population is $96$.
So we count $112$ monic polynomials sharing a factor, accounting for $672$ nonzero nonunits. $0$ is also a nonunit that we haven't counted so far, so the total is $673$ nonunits. Out of the entirity of $7^4=2401$ polynomials, you have $1728$ units.
If you know the Chinese Remainder theorem, this isn't hard to double-check. We'd have that $F_7[x]/(f(x))\cong F_7[x]/(x-2)\times F_7[x]/(x+2)\times F_7[x]/(x^2+2)\cong F_7\times F_7\times F_{49}$. THe units of the product ring are the product of the unit groups, so we'd expect $6\cdot 6\cdot 48$ units, which you see matches: $1728$.
Since that still only involved fields, I'll include one last example: suppose our $f(x)$ had been $(x-2)(x+2)^3$. The analysis along similar lines above says:
The totals is $106$ monic polynomials, or $636$ polynomials, therefore $1764$ units.
Double checking with the Chinese remainder theorem, we have that $F_7[x]/(f(x))\cong F_7[x]/(x-2)\times F_7[x]/(x+2)^3$, where the first factor is $F_7$ again, but the second factor is no longer a field. It's a local ring with maximal ideal $(x+2+(x+2)^3)$. As a local ring, its units are precisely the things not inside the maximal ideal: in this case, that ideal contains $7^2$ elements, leaving $7^3-7^2=294$ units behind. Then in the product, you guessed it, that accounts for $6\cdot 294=1764$ units.