Let $F$ be a field. Let $D\subseteq F$ be a subring of $F$ with the multiplicative identity. Put $D'=\{$ $ab^{-1}$ $:$ $a,b\in D$ , $b\neq 0$ $\}$. Then $D'$ is a subfield of $F$.
I'm in particular stuck with showing that every non-zero is a unit.
Here is my attempt so far: First, I will show that $D'$ is a subring of $F$. Indeed, $0\in D'$. Hence $D'$ is non-empty. Suppose $ab^{-1},cd^{-1}\in D'$. Then, $ab^{-1}-cd^{-1}=(ad-bc)(bd)^{-1}$. Since $D$ is an integral domain, $bd\neq 0$. Therefore the difference is in $D'$. Suppose $ab^{-1},cd^{-1}$ is in $D'$. Then, $(ab^{-1})(cd^{-1})=ac(bd)^{-1}\in D'$. Therefore $D'$ is a commutative subring. Note that $1\in D'$ and $1\neq 0$. Lastly, I will show that $D'$ is a division ring.
Let $ab^{-1}\in D’$ with $b\neq 0$, $a,b\in D$. If $a=0$ then $ab^{-1}=0$ and there is nothing to do. If $a\neq 0$, then isn’t $ba^{-1}\in D’$?