$F$ is a field. I consider that if Quotient ring of $F $by $N$ exists, then it would be zero ring or isomorphic to $F$.
$F$ is field, so it is ring. Suppose that $N$ is an ideal of ring$F$. And consider Quotient ring of $F$ by $N$.
If $N$ is zero set, then $F/N$ is equal to $F$ Otherwise N contains nonzero elements. And a Quotient Ring <$F/N$,+,•> can be a field.
That is my ideal. Please give me idea How to think more here.
Notice that the map $F \to F/N$ sending $a \mapsto a + N$ is in fact a linear map of $F$-vector spaces. This map is a surjection, so the rank-nullity theorem tells you that either $\dim_F(F/N) = 1$ or $\dim_F(F/N) = 0$. In the first case, the map is also injective and since the map is also a ring homomorphism, it is in fact a ring (field) isomorphism.
$F$ is a field, then $N=0$ or $N=F$. If $N=0\Rightarrow F/N=F$ and if $N=F\Rightarrow F/F=0$
Concerning your title question: since $F$ is a field, it is also a domain, so that $F$ embeds into its quotient ring $Q(F)$. This says in particular, that with $F\neq 0$ we have $Q(F)\neq 0$. The construction of a quotient field shows of course that $Q(F)\cong F$.
Suppose $F$ is a field, and $N$ is an ideal of $F$.
If $N = \{0_F\}$, clearly $F/N \cong F$.
Otherwise, we have $a \neq 0 \in N$, and thus $1_F = (a^{-1})a \in N$, and thus for any $b \in F$, we have $b = b(1_F) \in N$, and thus $N = F$, so that $F/N \cong \{0_F\}$ (a zero ring).