Denote $J_n:=\mathbb{C}(xy^n+x^ny)$, $K_n:=\mathbb{C}(x+y^n,x^n+y)$ and $L:=\mathbb{C}(x,y)$.
Notice that $J_n(x):=\mathbb{C}(xy^n+x^ny,x) \subsetneq L$ and $J_n(y):=\mathbb{C}(xy^n+x^ny,y) \subsetneq L$, while $K_n(x):=\mathbb{C}(x+y^n,x^n+y,x) = L$ and $K_n(y):=\mathbb{C}(x+y^n,x^n+y,y) = L$.
How to compute the degrees of the field extensions $[L:J_n]$ and $[L:K_n]$?
It seems appropriate to begin with the case $n=1$: $J_1:=\mathbb{C}(xy+xy)=\mathbb{C}(xy)$, $K_1:=\mathbb{C}(x+y,x+y)=\mathbb{C}(x+y)$
It seems that $[\mathbb{C}(x,y):\mathbb{C}(xy)]=\infty$ and $[\mathbb{C}(x,y):\mathbb{C}(x+y)]=\infty$ because $x$ is transcendental over $\mathbb{C}(xy)$ and $\mathbb{C}(x+y)$. Similarly for any $n \in \mathbb{N}-\{0\}$.
Notice that for $n=0$, $J_0:=\mathbb{C}(x+y)$, $K_0:=\mathbb{C}(x+1,1+y)=\mathbb{C}(x,y)$, so $[\mathbb{C}(x,y):K_0]=1$.
I first thought that $[\mathbb{C}(x,y):K_n]=n+1$ but even for $n=1$ I have not found a polynomial $P(t) \in \mathbb{C}(x+y)[t]$ of degree two such that $P(x)=0$ (or $P(y)=0$). The problem is that $xy \notin \mathbb{C}(x+y)$, so we are not able to take $P(t)=t^2-(x+y)t+xy$ as a minimal polynomial for $x$, for example.
Perhaps if we consider fields similar to $\mathbb{C}(x+y,xy)$ (for which $[L:\mathbb{C}(x+y,xy)]=2$), then we can obtain finite degrees field extensions, for example: $\mathbb{C}(xy^2+x^2y,x+y^3,x^3+y)$ (still seems of infinite degree). It seems that some symmetry is required.
Any help is welcome, thank you!
For $n\ge 2$, $[L,K_n]$ is finite and computable. First, note that $K_n(x)=L$ because $y=(x^n+y)-x^n$.
We have the relation $$(x^n+y-x^n)^n=x+y^n-x.$$
Let $u=x+y^n, v=x^n+y$. Then, $$(-x^n+v)^n+x-u=0.$$ This shows that $x$ is algebraic over $K_n$ for $n\ge 2$.
In fact, the LHS is irreducible as a polynomial in $x$ with coefficients in $K_n$: Because $K_n$ is isomorphic to the field of rational functions in two variables, we can apply Gauss lemma and it suffices to show that it is irreducible in $\mathbb{C}[u,v,x]$. It is monic and of degree one in $u$ so we are done. As a byproduct we obtain $[L,K_n]=n^n$. This is compatible with the case $n=0$ ($0^0=1$).