I would like to show that the rational polynomial function $f:\mathbb{Q}^2\to\mathbb{Q}$ defined as $f(x,y)=xy(x+y)$ isn't surjective. To do this, I tried proving that the algebraic plane curve $$K:xy(x+y)=1$$ has no rational points so that $1\not\in f[\mathbb{Q}^2]$. I then considered the following elliptic curve $$E:v^2-v=u^3$$ which, according to the LMFDB, has the three rational points $E(\mathbb{Q})=\{\mathcal{O},(0,0),(0,1)\}\cong\mathbb{Z}/3\mathbb{Z}$. Now, the function $\psi:K(\mathbb{Q})\to E(\mathbb{Q})$ defined as $\psi(x,y)=\left(- \frac{1 }{x+y},\frac{y}{x+y}\right)$ is well defined since $x+y$ isn't annihilated on $K$ and $$v^2-v=\left(\frac{y}{x+y}\right)^2-\frac{y}{x+y}=\frac{y^2-y(x+y)}{(x+y)^2}=-\frac{xy}{(x+y)^2}=-\frac{1}{(x+y)^3}=u^3$$ with $(u,v)=\psi(x,y)$. But this means that $K$ has no rational points since $-\frac{1}{x+y}\neq0\Rightarrow \psi(x,y)\not\in E(\mathbb{Q}).\ \square$
However, this seems too much as it relies on theory of elliptic curves computing the Mordell-Weil group of $E$. Maybe there's a non-constructive approach which proves that some $q\in\mathbb{Q}$ isn't in the image of $f$ without specifying it. Or with elementary number theory employing, for example, modular arithmetic and power residues. So my question then is:
Q: Is there an easier approach on proving that $f$ is not surjective?
With elementary reasoning we can reduce to Fermat's Last Theorem for $n=3$, which in turn does have an elementary proof (see e.g. Euler's proof).
That is, first write $x=\frac{r_1}{s_1}$, $y=\frac{r_2}{s_2}$ with $\gcd(r_i,s_i)=1$, then let $r=\gcd(r_1,r_2)$, $s=\gcd(s_1,s_2)$, so we may write
$x=\frac{p_1r}{q_1s}$, $y=\frac{p_2r}{q_2s}$ with $\gcd(p_1,p_2)=\gcd(q_1,q_2)=1$.
We assume only $q_1$ and $q_2$ may be negative.
We claim that we can never have $xy(x+y)=1$. To see this, suppose otherwise and expand to obtain
$$1=\frac{r^3(p_1q_2+q_1p_2)p_1p_2}{s^3q_1^2q_2^2}.$$
From the fact that $r$ is positive and coprime to $s$, $q_1$, and $q_2$, we have $r=1$. Moreover, since $q_1$ is coprime to $p_1$ and $q_2$, it is coprime to $(p_1q_2+q_1p_2)$, and therefore we have $q_1^2|p_2$, and similarly $q_2^2|p_1$.
Now writing $p_1=k_1q_2^2$, $p_2=k_2q_1^2$ we have
$$1=\frac{(k_1q_2^3+k_2q_1^3)k_1q_2^2k_2q_1^2}{s^3q_1^2q_2^2}=\frac{(k_1q_2^3+k_2q_1^3)k_1k_2}{s^3}.$$
Since $p_i$ and $s$ are coprime and positive we then have $k_1=k_2=1$, so that
$$1=\frac{q_2^3+q_1^3}{s^3}$$
the desired contradiction to Fermat's Last Theorem for $n=3$.