Degree of $F(x+y)$ over $F=k(x^3,y^3)$ in characteristic $2$

Question

Let $k$ be a field and let $x$ and $y$ be indeterminates. Consider $F=k(x^3,y^3)$. One can easily show that $F(x+y)$ is a degree $3$ extension of $F$ when $k$ has characteristic $3$. I was also able to show that $F(x+y)=F(x,y)$, a degree $9$ extension, when the characteristic of $k$ is neither $2$ nor $3$ by expressing $x$ and $y$ as rational functions in terms of $x+y$ with coefficients in $F$. But my expression required dividing by $2$ (and $3$), so I am left to wonder

what is the degree of $F(x+y)$ over $F=k(x^3,y^3)$ when $k$ has characteristic $2$?

Edit: For those curious, here was how I wrote $x$ and $y$ as a rational function of $x+y$ over $F$:

First, we may express $x-y$ as $$x-y=\frac{(x-y)(x^2+xy+y^2)}{x^2+xy+y^2}=\frac{x^3-y^3}{(x+y)^2-\frac{(x+y)^3-x^3-y^3}{3(x+y)}}\in F(x+y).$$

Once we have both $x+y$ and $x-y$, we can write $$x=\frac{1}{2}(x+y)+\frac{1}{2}(x-y),$$ $$y=\frac{1}{2}(x+y)-\frac{1}{2}(x-y).$$ But of course, you can see how these last expressions don't work in characteristic $2$. Thank you to Jyrki Lahtonen for commenting a solution that works even in characteristic $2$!

Answer 1

$F(x,y)/F$ is separable of degree $9$

The $F$-conjugates of $x,y$ are $\zeta_3^n x,\zeta_3^m y$

So the $F$-conjugates of $x+y$ are $\zeta_3^n x+\zeta_3^m y$

$x+y$ has $9$-distinct $F$-conjugates so $[F(x+y):F]=9$.