I believe it is $\delta(xy)=\delta(x)+\delta(y)$.
Let $F=\{x:\delta(x)=0\}$. We will prove that $F$ is a field.
Let $x,y\in F$, then: $ \delta(xy)=\delta(x)+\delta(y)=0, \delta(x+y)\leq \text{max}(\delta(x),\delta(y))=0$, so $x+y,xy \in F$.
$1=ax+r,\delta(r)<\delta(x)=0$, or $r=0$ implies $r=0$ since $\delta(r)$ is not negative. $1=ax$ implies that $a=a^2x$ and $\delta(a)=\delta(a^2)\delta(x)=0$ implies that $\delta(a)=0, a\in F$.
If $F=R$, then R is a field, and we are done. Suppose now $F\neq R$, i.e. there exists $u:\delta(u)>0$.
Let $x\in R$, having the smallest positive degree amongst all numbers with positive degree, i.e. $\delta(x)>0$ and $\forall y\in R$, $\delta(y)>0 \Rightarrow \delta(x)\leq\delta(y)$, We will prove that $R=F[x]$.
Proof. Recursive on $\delta(y)$ suppose true for $n$, let $\delta(y)=n+1$, we can write $y=zx+b\delta(b)<\delta(x)$ implies that $\delta(b)=0$ and $b\in F$.
$\delta(zb)=\delta(y-b)\leq \delta(y)$. We deduce that $\delta(zb)=\delta(z)+\delta(b)$ and $\delta(z)<n+1$ we deduce that $z\in F[x]$ and $y\in F[x]$.