= Solution
The <Hessian criterion for flexes of a plane cubic> applied to the defining <polynomial> gives
$$
\det\begin{pmatrix}6X&0&0\\0&6Y&0\\0&0&6dZ\end{pmatrix}=216dXYZ.
$$
Consequently the <inflection points of a plane cubic> on this curve are exactly its intersections with the three coordinate lines. Choose $r\in\overline{\mathbb Q}$ with $r^3=d$, and a primitive cube <root of unity> $\zeta$. The nine distinct <flexes of a diagonal cubic> are
$$
\boxed{(1:-\zeta^j:0),\quad(0:-r\zeta^j:1),\quad(-r\zeta^j:0:1),\qquad j=0,1,2.}
$$
With the chosen flex as identity, these points are precisely $E_d[3]$. A tangent cutting the <Weil divisor> $3P$ gives $3(P-O)=0$; conversely $3(P-O)=0$ makes the hyperplane <Weil divisor> equivalent to $3P$, which supplies a tangent with triple contact. This is the <three-torsion points are flexes of a plane cubic> correspondence.
On $Z=0$ only $O$ is rational, since the other cube <roots of unity> are not rational. On each other coordinate line a <rational point> exists exactly when $d$ is a rational cube, and then exactly one does. Therefore
$$
\boxed{E_d(\mathbb Q)[3]\cong\begin{cases}\mathbb Z/3\mathbb Z,&d\in(\mathbb Q^\times)^3,\\0,&\text{otherwise}.\end{cases}}
$$
In particular the rational <torsion subgroup> cannot contain two independent points of order $3$.
One can also determine the whole <rational torsion of a diagonal cubic>. First it is killed by $6$. To see this, use the <Weierstrass equation of an elliptic curve> from (iii). At <primes> $\ell\equiv2\pmod3$ outside the finite set containing $2,3$ and the prime divisors of the numerator and denominator of $d$, the cube map permutes $\mathbb F_\ell$, so
$$
\#E_d(\mathbb F_\ell)=1+\sum_{x\in\mathbb F_\ell}\bigl(1+\chi(x^3-432d^2)\bigr)=\ell+1,
$$
where the quadratic character is extended by $\chi(0)=0$, and the character sum vanishes after substituting $t=x^3-432d^2$. <Reduction of torsion points on an elliptic curve> injects the prime-to-$\ell$ part of rational torsion into this <finite group>. This injection follows because multiplication by an <integer> <prime> to $\ell$ is invertible on the <formal group of an elliptic curve>, and hence its reduction kernel has no such torsion.
For any <prime> $q>3$, choose a good $\ell\equiv2\pmod3$ with $\ell\equiv1\pmod q$; the <Dirichlet theorem on primes in arithmetic progressions> and the <Chinese remainder theorem> supply such <primes> outside the finite bad <set>. Then $q\nmid\ell+1$, excluding $q$-torsion. Similarly good $\ell\equiv5\pmod{12}$ exclude order $4$, and good $\ell\equiv11\pmod{18}$ exclude order $9$. Thus every rational torsion point has order dividing $6$.
The inversion formula in (i) shows that the only possible nonidentity <rational point> of order $2$ has $X=Y$ and $Z\ne0$. It exists precisely when $(X/Z)^3=-d/2$, that is, when $d/2$ is a rational cube. There is at most one such point. This condition and $d$ being a cube cannot hold together, since $2$ is not a rational cube. Combining it with the three-torsion calculation gives the full answer:
$$
\boxed{E_d(\mathbb Q)_{\mathrm{tors}}\cong\begin{cases}
\mathbb Z/3\mathbb Z,&d\in(\mathbb Q^\times)^3,\\
\mathbb Z/2\mathbb Z,&d/2\in(\mathbb Q^\times)^3,\\
0,&\text{otherwise}.
\end{cases}}
$$
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