 Solution

For $x=u/v\in \mathbb{Q}$ in lowest terms, the naive height on the projective line isHomogenize the coprime numerator and denominator of $\xi$ to degree $d$. The triangle inequality gives the upper estimate $H(\xi (x))\le c_{2}H(x{)}^d$. 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 $H(x{)}^d\le CH(\xi (x))$, which is the lower estimate with $c_1=C^{-1}$.

Now write $x=u/v$ in lowest terms and putSince $\gcd (N,v)=1$, the equation $y^2=N/v^3$ givesClearly $M\le \gamma H(x{)}^3$ for $\gamma =1+|a|+|b|$. Homogenizing the supplied polynomial identity givesIts coefficient sum is at most ${\gamma }^2$, so $|u{|}^5\le {\gamma }^{2}H(x{)}^{2}M$. The same lower bound is immediate from $|v{|}^3\le M$ when $|v|=H(x)$. Thus