= Solution
For <rank-one two-isogeny descent with parameters three and minus three>, use the standard <two-isogeny rank formula>. For $C:y^2=x(x^2+ax+b)$ and its dual curve $D:y^2=x(x^2-2ax+a^2-4b)$, let $\alpha_C,\alpha_D$ be the abscissa <square class>[square-class] maps, with their usual values at $(0,0)$ and the identity. If $r_C$ is the <rank of an elliptic curve>[Mordell–Weil rank], then
$$
\boxed{2^{r_C+2}=|\operatorname{im}\alpha_C|\,|\operatorname{im}\alpha_D|.}
$$
For the present curve $a=3,b=-3$, its possible classes are $1,-1,3,-3$, and the identity and $(0,0)$ supply $1,-3$. Exclude the class three as follows. A point in that class has $x=3(U/V)^2$, with coprime <integers> $U,V$, and substitution in the curve gives the necessary quartic equation
$$
N^2=3U^4+3U^2V^2-V^4
$$
with <integer> $N$. Modulo three this forces $3\mid V,N$. Coprimality then makes $U$ a <unit in a ring>[unit] modulo three, and modulo nine the right side is $3U^4\equiv3$, while $N^2\equiv0$, a contradiction. The class minus one is excluded as well, since multiplying it by the known class minus three would give three. Therefore the first image has size two.
The dual curve is $D:y^2=x(x^2-6x+21)$. Its quadratic factor is $(x-3)^2+12>0$ on the <real line>, so every nonzero rational abscissa is positive. By (i) its image is contained in $\{1,3,7,21\}$. The point $(0,0)$ supplies 21 and the point $(3,6)$ supplies three. As the <square class>[square-class] map is a homomorphism, their product supplies seven. All four occur. Hence the descent formula gives $2^{r_C+2}=2\cdot4=8$, so
$$
\boxed{\operatorname{rank}C(\mathbb Q)=1.}
$$
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