From the previous parts I've proven that
$$D:F[x]\to F[x]$$
is an additive group homomorphism on addition for $F[x]$ and not a ring homomorphism because the multiplication does not hold.
OK, so I do not understand why the kernel of $D$ is $F$. Can someone please explain that intuitively to me? Also is the image $D(F[x])$, $F[x]$ because its mapping into itself? If you can clarify my reasoning for this that would be great.
Formal differentiation takes a polynomial
$p(x)=a_Nx^N+\ldots +a_1x+a_0$ and produces
$$Dp(x)= Na_Nx^{N-1}+(N-1)a_{N-1}x^{N-2}+\ldots + a_2x+a_1$$
Then in order for this to be the zero polynomial--since the field has characteristic $0$--it must be that $a_N, a_{N-1},\ldots, a_1=0$, otherwise the largest $k$ such that $a_k\ne 0$ gives a non-zero leading term $ka_kx^{k-1}$ for $Dp(x)$. This implies that kernel elements are polynomials of the form $p(x)=a_0$ is a constant. But by definition this means that the kernel is polynomials of degree $0$, i.e. $F$.
To see that $D$ is surjective, we just note that we can use the formal integral to produce a polynomial $P(x)$ such that $DP(x)=p(x)$. Namely
$$P(x)={1\over N+1}a_Nx^{N+1}+{1\over N}a_{N-1}x^{N}+\ldots+{1\over 2}a_1x^2+a_0x+C$$
for any $C\in F$.