= Solution
Here $a=1$, $b=-2$ and the two-isogenous curve is $D:v^2=u(u^2-2u+9)$. Let $\delta_C$ and $\delta_D$ be the first-coordinate square-class maps, using the coefficient $b$ or $b_1$ at the point $(0,0)$. The same chord argument shows that both maps are homomorphisms.
For an equation $y^2=x(x^2+ax+b)$ with <integer> coefficients, its first-coordinate <square class> has a square-free representative dividing $b$, including a possible sign. Indeed, at a prime not dividing $b$, a negative valuation of $x$ must be even because $x^3$ dominates; a positive valuation must be even because $x^2+ax+b$ is a unit. Thus only primes dividing $b$ can occur to odd order. For $C$, the possibilities are $1,-1,2,-2$. The classes $1,-2$ occur at the identity and $(0,0)$. If class two occurred, write $x=2M^2/e^2$ with coprime <integers> $M,e$. Substituting in the equation would give a rational square, necessarily an <integer> square,
$$
N^2=2M^4+M^2e^2-e^4.
$$
Modulo three, the right side is two whenever $M,e$ are not both divisible by three: this holds whether just one or both are units. That is impossible. Since the image is a subgroup containing $-2$, exclusion of two also excludes minus one. Therefore $|\delta_C(C(\mathbb Q))|=2$.
For $D$, the quadratic $u^2-2u+9=(u-1)^2+8$ is positive, so every nonzero first coordinate is positive. Its <square class> can only be one or three. Both occur, with three supplied by $(3,6)$, so $|\delta_D(D(\mathbb Q))|=2$.
Let $\psi:D\to C$ be the dual two-isogeny, with $\psi\phi=[2]$. The kernel identities of part (ii), applied to both curves and the usual coordinate scaling, give $[C:\psi D]=2$ and $[D:\phi C]=2$. Moreover $\ker\psi=\{O,T_1\}$ lies in $\phi C$, since $b_1=9$ is a square. Thus $[\psi D:2C]=[D:\phi C]=2$, and
$$
[C(\mathbb Q):2C(\mathbb Q)]=4.
$$
This is a <rank-zero two-isogeny descent for roots zero one and minus two>. The rational two-torsion has four points, since the cubic factors as $x(x-1)(x+2)$. By the <Mordell-Weil theorem>, the last index is $2^r\,|C(\mathbb Q)[2]|=2^{r+2}$. Therefore
$$
\boxed{\operatorname{rank}C(\mathbb Q)=0.}
$$
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