5 added 31 characters in body

I think that the answer is no.

Here is a somewhat related problem. Every curve curve $C$ of genus >0, has meromorphic functions $x,y$ on it which are not related by any equation of the form $f(x)=g(y)$. I denote by $(x)=(x)^+-(x)^-$ the principal divisor of an element $x$, zeros minus poles. If $x$ and $y$ are two elemets of this filed, the field of meromorphic functions on $C$, related by $f(x)=g(y)$, where $f,g$ are ratonal functions, then the divisors of poles of $x$ and $y$ are related as follows: $$m(x)^-\sim n(y)^-,$$ where $\sim$ means the usual equivalence of divisors. (Two dividors $d$ and $e$ are equivalent if $d=e+(z)$). And $m,n$ are degrees of $f,g$.

Now the factor of the set of all divisors over this equivalent equation is a torus of dimension $g$ ($g$ is the genus of $C$). We only need the fact that it is uncountable for $g>0$. So we can always find incommensurable divisors of the form $(x)^-$ and $(y)^-$. These $x$ and $y$ are related by some polynomial relation $F(x,y)=0$, but cannot be related by an equation of the form $f(x)=g(y)$.

This solution was explained me by Drinfeld in 1980 when I asked him more general question: Can every algebraic relation $F(x,y)=0$ be obtained from a chain $x=x_1,x_2,x_3,\ldots,x_n=y$ where $x_i$ and $x_{i+1}$ are related by $f_i(x_i)=f_{i+1}(x_{i+1})$, with some rational functions $f_i$, by elimination of $x_2,...x_{n-1}$? The answer is no, for the same reason).

4 deleted 498 characters in body

The answer is no. I consider everything over complex numbers (the only numbers

I know; let other people explain what happens in other fleds).

EDIT:

1. Let $d$ be the maximal degree of $d$ and $g$. Then think that the equations $f(x)=g(y)$ depend on $4d+2$ parameters. The curve $(f(x)-f(y))/(x-y)$ answer is of degree $d-1$ and genericaly non-singularno.Therefore its genus

Here is $g=(d-2)(d-3)/2$. But generic curves of genus $g$, depend on $g$ parameters. This implies that for high genus, a generic curve cannot be represented in this way.

2. A somewhat related problem. Every curve curve $C$ of genus >0, has meromorphic functions $x,y$ on it which are not related by any equation of the form $f(x)=g(y)$. I denote by $(x)=(x)^+-(x)^-$ the principal divisor of an element $x$, zeros minus poles. If $x$ and $y$ are two elemets of this filed, related by $f(x)=g(y)$, where $f,g$ are ratonal functions, then the divisors of poles of $x$ and $y$ are related as follows: $$m(x)^-\sim n(y)^-,$$ where $\sim$ means the usual equivalence of divisors. (Two dividors $d$ and $e$ are equivalent if $d=e+(z)$). And $m,n$ are degrees of $f,g$.

Now the factor of the set of all divisors over this equivalent equation is a torus of dimension $g$ ($g$ is the genus of $C$). We only need the fact that it is uncountable for $g>0$. So we can always find incommensurable divisors of the form $(x)^-$ and $(y)^-$. These $x$ and $y$ are related by some polynomial relation $F(x,y)=0$, but cannot be related by an equation of the form $f(x)=g(y)$.

This solution was explained me by Drinfeld in 1980 when I asked him more general question: Can every algebraic relation $F(x,y)=0$ be obtained from a chain $x=x_1,x_2,x_3,\ldots,x_n=y$ where $x_i$ and $x_{i+1}$ are related by $f_i(x_i)=f_{i+1}(x_{i+1})$, with some rational functions $f_i$, by elimination of $x_2,...x_{n-1}$? The answer is no, for the same reason).

3 added 73 characters in body

The answer is no. I consider everything over complex numbers (the only numbers I know; let other people explain what happens in other fleds).

EDIT:

1. Let $d$ be the maximal degree of $d$ and $g$. Then the equations $f(x)=g(y)$ depend on $4d+2$ parameters, while generic curves . The curve $(f(x)-f(y))/(x-y)$ is of degree $d$ have d-1$and genericaly non-singular. Therefore its genus is$(d-1)(d-2)/2$, thus g=(d-2)(d-3)/2$. But generic curves of genus $g$, depend in on $3(d-1)(d-2)/2-3$ g$parameters. This implies that for high genus, a generic curve cannot be represented in this way. 2. A somewhat related problem. Every curve curve$C$of genus >0, has meromorphic functions$x,y$on it which are not related by any equation of the form$f(x)=g(y)$. I denote by$(x)=(x)^+-(x)^-$the principal divisor of an element$x$, zeros minus poles. If$x$and$y$are two elemets of this filed, related by$f(x)=g(y)$, where$f,g$are ratonal functions, then the divisors of poles of$x$and$y$are related as follows: $$m(x)^-\sim n(y)^-,$$ where$\sim$means the usual equivalence of divisors. (Two dividors$d$and$e$are equivalent if$d=e+(z)$). And$m,n$are degrees of$f,g$. Now the factor of the set of all divisors over this equivalent equation is a torus of dimension$g$($g$is the genus of$C$). We only need the fact that it is uncountable for$g>0$. So we can always find incommensurable divisors of the form$(x)^-$and$(y)^-$. These$x$and$y$are related by some polynomial relation$F(x,y)=0$, but cannot be related by an equation of the form$f(x)=g(y)$. This solution was explained me by Drinfeld in 1980 when I asked him more general question: Can every algebraic relation$F(x,y)=0$be obtained from a chain$x=x_1,x_2,x_3,\ldots,x_n=y$where$x_i$and$x_{i+1}$are related by$f_i(x_i)=f_{i+1}(x_{i+1})$, with some rational functions$f_i$, by elimination of$x_2,...x_{n-1}\$? The answer is no, for the same reason).

2 added 410 characters in body; deleted 1 characters in body
1