Determine if $E(\mathbb{Q})$ is finite or infinite. It is given the following algebraic curve:
$$ZY^2=X^3+3XZ$$
I want to find the group of rational points of finite order $E(\mathbb{Q})_{\text{torsion}}$ and to determine if $E(\mathbb{Q})$ is finite or infinite.
I thought the corresponding curve in affine coordinates is: $Y^2=X^3+3X$ with $a=3, b=0$.
I found that $E(\mathbb{Q})_{\text{torsion}}=\{ (0,0), [0,1,0]\}$.
Then to determine if $E(\mathbb{Q})$ is finite or not, I did the following:
$$E|_{\overline{Q}}: Y^2=X^3-6X$$
$$2^r=\frac{|a \Gamma| |\overline{a} \overline{\Gamma}|}{4}$$
where $r$ is the rank of the curve.
$a \Gamma=\{ 1 \mathbb{Q}^{{\star}^2}, b \mathbb{Q}^{{\star}^2} \} \cup \{ b_1 \mathbb{Q}^{{{\star}^2}} | b_1 \mid b \text{ and the diophantine equation } Z^2=b_1X^4+aXY+b_2Y^4 \text{ has a solution}\} $
In our case $b=0$ and thus the set $\{ b_1 \mathbb{Q}^{{{\star}^2}} | b_1 \mid b \text{ and the diophantine equation } Z^2=b_1X^4+aXY+b_2Y^4 \text{ has a solution}\} $ is finfinite, so we deduce that $E(\mathbb{Q})$ is infinite.
Could you tell me if it is  right?

You have the torsion group right and $E(\mathbb{Q})$ is infinite. The rank is $1$ with a generator being $P = (1,2)$.
I think your reasoning for $E(\mathbb{Q})$ being infinite is flawed. You are implicitly using the method that originate from Tate that exploits the homomorphism $ a : E \rightarrow \mathbb{Q}^* / {\mathbb{Q}^*}^2$. 
Now, the $2$-isogenous curve to $E$ should be $ \bar{E}: y^2 = x^3-12x$ and because $3$ is a prime we have that $|a(E)| = 2$. To compute $|\bar{a}(\bar{E})|$ we have to consider the solutions of the following equations
$$X^4-12Y^4=Z^2$$
$$2X^4-6Y^4=Z^2$$
$$3X^4-4Y^4=Z^2$$
$$-X^4+12Y^4=Z^2$$
$$-2X^4+6Y^4=Z^2$$
$$-3X^4+4Y^4=Z^2$$
$(X,Y,Z)=(2,1,2)$ is a solution to the first equation, $(X,Y,Z)=(1,1,2)$ and $(X,Y,Z)=(1,1,1)$ are solutions to the last two equations respectively. Hence, $3 \leq |\bar{a}(\bar{E})| \leq 6$, and so $|\bar{a}(\bar{E})| = 4$ for the rank to be integral.
Hopefully I got all the details right.
