Hint: This is primarily a counterexample that shows the kernel to be non-trivial, but it also serves as a strong hint. Let $f(x) = x^2 - 3x + 2$. Then $\varphi(f(x)) = f(1) = 1 - 3 + 2 = 0 \implies f(x) \in \ker \varphi$.
Solution:
\begin{align*}
\ker \varphi & = \{\, f(x) \in \mathbb R[x] \mid \varphi(f(x)) = 0 \in \mathbb R \,\}\\
& = \{\, f(x) \in \mathbb R[x] \mid f(1) = 0 \,\}\\
& = \{\, f(x) \in \mathbb R[x] \mid (x - 1)\ \text{is a factor of}\ f(x) \,\}\\
& = \{\, (x -1)g(x) \mid g(x) \in \mathbb R[x] \,\}\\
& = \langle x - 1 \rangle
\end{align*}
That is, $\ker \varphi$ is the ideal generated by $x - 1$.
In a word, as I said in a comment, for there to be such a ring morphism there would have to be a transcendence-degree-two subfield of the field $\Bbb Q(t)$, which has transcendence degree one. Impossible, because the absolute transcendence degree of $\Bbb Q$ is zero. This argument fails when the base has infinite transcendence degree over $\Bbb Q$, like $\Bbb C$.
Be that as it may, giving a direct proof is not hard, though it can be tiresomely long (especially when a wordy geezer is at the keyboard). I’ll appreciate suggestions for shortening what appears below.
As you have observed, the ring morphism in question has to be one-to-one, because the domain is a field, so you take the images $a(t),b(t)$ of $x,y$ respectively in $\kappa=\Bbb Q(t)$ and hope to find a nonzero $\Bbb Q$-polynomial $F(X,Y)$ such that $F(a,b)=0$. Since $F(x,y)\ne0$, there’s your contradiction.
The proof is in two parts, the easy and then the harder (at least the longer). First part is, having chosen $a(t)$ as a starting point, to show that $t$ is algebraic over $\kappa=\Bbb Q(a)$. Second part is to take $b(t)$, now known to be algebraic over $\kappa$ (because everything in the big field is now algebraic over $\kappa$), and take its minimal $\kappa$-polynomial and convert this to a $\Bbb Q$-polynomial $F$ of the desired type.
So much for the program. Now to expand it to the tiresome totality.
Let $a(t)=g(t)/h(t)$ where $g$ and $h$ are $\Bbb Q$-polynomials. Now form $a\!\cdot\! h(T)-g(T)\in\Bbb Q(a)[T]$, which you see is a polynomial over $\kappa$ that vanishes at $T=t$, so that $t$ is algebraic over $\kappa$. (This is the appearance of transcendence-degree one in the argument.)
That was the quick and easy part. Now $b$ is also algebraic over $\kappa$, so
that it satisfies a monic $\kappa$-polynomial
$$
\Phi(Z)=Z^m+c_{m-1}Z^{m-1} +\cdots+c_1Z+c_0=Z^m+\frac{g_{m-1}}{h_{m-1}}Z^{m-1}
+\cdots+\frac{g_1}{h_1}Z+\frac{g_0}{h_0}\,.
$$
Here, the $g_i$ and the $h_i$ are in $\Bbb Q[a]$. When you multiply the displayed minimal polynomial for $b$ by the product of all the $h_i$, call it $H(a)$, you get the polynomial
$$
H(a)\Phi(Z)=H(a)Z^m+\gamma_{m-1}(a)Z^{m-1}+\cdots+\gamma_1(a)Z+\gamma_0(a)\,,
$$
where each $\gamma_i=g_i(a)\!\cdot\!\bigl(H(a)/h_i(a)\bigr)$, an element of $\Bbb Q[a]$. Make the substitution $Z\mapsto b(t)$ and get zero. This is your $\Bbb Q$-polynomial in two variables vanishing at $(a,b)$.
Best Answer
Observe that in general, if $d = \gcd(a,b)$, then $x^{b/d} - y^{a/d} \in \ker(\phi)$, so $\langle x^{b/d} - y^{a/d} \rangle \subseteq \ker(\phi)$.
To show the reverse inclusion, notice that by treating $\mathbb{F}[x,y]$ as $\mathbb{F}[y][x]$, using the division algorithm we can write any $f \in \mathbb{F}[x,y]$ as $q(x,y) (x^{b/d} - y^{a/d}) + r(x,y)$ where each term in the remainder $r(x,y)$ has $x$-degree less than $b/d$. Now, $\phi(f) = \phi(r)$. On the other hand, each possible monomial $x^m y^n$ with $0 \le m < b/d$ maps to a distinct monomial $z^{am + bn}$ (see below for a proof), so if $f \in \ker(\phi)$ then $\phi(r) = 0$ implies $r = 0$.
So, all we have left to show is the easy purely number theoretic result:
To show this, suppose $am + bn = am' + bn'$; then $(a/d) (m-m') = (b/d) (n'-n)$. Since $\gcd(a/d, b/d) = 1$, this implies that $m \equiv m' \pmod{b/d}$; but due to the restriction on the range of possible values of $m,m'$, that further implies that $m=m'$ and therefore $n=n'$ also.
(Note that this argument assumes that $0 \notin \mathbb{N}$ in your convention, so that $a \ne 0$ and $b \ne 0$. If either or both of $a,b=0$, then those special cases should be easier to work with.)