= Solution
For $x=u/v\in\mathbb Q$ in lowest terms, the <naive height on the projective line> is
$$
H(x)=\max\{|u|,|v|\}.
$$
Homogenize the coprime numerator and denominator of $\xi$ to degree $d$. The triangle inequality gives the upper estimate $H(\xi(x))\leq c_2H(x)^d$. Since the two homogenized forms have no common projective zero, their <resultant> is nonzero, and the Bézout identities for the resultant express fixed multiples of $u^{2d-1}$ and $v^{2d-1}$ as combinations of their values with coefficients of degree $d-1$. After cancellation this gives $H(x)^d\leq C H(\xi(x))$, which is the lower estimate with $c_1=C^{-1}$.
Now write $x=u/v$ in lowest terms and put
$$
N=u^3+auv^2+bv^3.
$$
Since $\gcd(N,v)=1$, the equation $y^2=N/v^3$ gives
$$
H(y)^2=\max\{|N|,|v|^3\}=:M.
$$
Clearly $M\leq\gamma H(x)^3$ for $\gamma=1+|a|+|b|$. Homogenizing the supplied polynomial identity gives
$$
u^5=(u^2-av^2)N-(bu^2-a^2uv-abv^2)v^3.
$$
Its coefficient sum is at most $\gamma^2$, so $|u|^5\leq\gamma^2H(x)^2M$. The same lower bound is immediate from $|v|^3\leq M$ when $|v|=H(x)$. Thus
$$
\gamma^{-2}H(x)^3\leq H(y)^2\leq\gamma H(x)^3.
$$
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