Let $E,E'$ be two elliptic curves over a field $k$ of characteristic $\neq 2, 3$. Assume that $E,E'$ are given by short Weierstrass equations, and let $f : E \to E'$ be a morphism given by $$f(x,y) = (f_1(x, y), f_2(x, y))$$
Question: is it true that $f_1(x, y) \in k(x)$ ? That is, does the first coordinate of $f$ only depend on $x$? It seems to be the case on all examples I've checked, but I don't see why it should be the case in general.
Remark: it is not true if $E' = E$ is not given by a Weierstrass equation, for instance $E : X^3 + Y^3 = Z^3$ is a smooth projective plane cubic curve so it has genus 1 (and has $\Bbb Q$-points), so it is an elliptic curve, but $f :E \to E$ defined by $(x, y) \mapsto (y, x)$ does not have its first component depending only on $x$.
So somehow, we should use the double cover $\pi : E \to \mathbb P^1, (x,y) \mapsto x$ (of degree 2) given by the Weierstrass equation.
Since $f$ is an isogeny, it is a group homomorphism, so $f(-P) = -f(P)$, or in other words $f(x_P,-y_P) = (f_1(x_P,-y_P), f_2(x_P,-y_P)) = (f_1(x_P,y_P), -f_2(x_P,y_P))$. In particular, $f_1(x_P,-y_P) = f_1(x_P,y_P)$. Therefore $f_1(x,y)$, as a polynomial (it's actually a rational function, but "polynomial" is close enough for our purposes), is an even function with respect to $y$. This means every occurrence of $y$ in $f_1(x,y)$ must be of even degree. But every term of the form $y^{2k}$ is purely a function of $x$, because $y^{2k} = (x^3+ax+b)^k$.