The $4 x^3 = y^2 +27$ solutions. Is it true that if 
$$4 x^3 = y^2 +27$$
then 
$$y=3k$$
I tried the Fermat theorem for $x^3$, but seems I am missing something.
EDITED:
$x$ and $y$ are integers.

You are correct that you could use Fermat's theorem for $x^3+y^3=z^3$.
Starting with your elliptic curve
$$4x^3=y^2+27$$
we borrow Yong Hao Ng's answer by multiplying it by $16$ to get
$$(4x)^3=(4y)^2+432$$

$$X^3-432=Y^2\tag1$$

This is a special case of a well-known family. The problem of finding two rational cubes equal to $N$
$$p^3+q^3 = N$$
or,
$$\left(\frac{36N+v}{6u}\right)^3+\left(\frac{36N-v}{6u}\right)^3 =N$$ 
simplifies to

$$u^3-432N^2 =v^2\tag2$$

Thus, your curve is just the case $N=1$. Of course, by Fermat's Last Theorem, there are no non-trivial rational solutions to,
$$p^3 + q^3 =1$$
so the only rational point on $(1)$ is its torsion point $X=12$, and no other.
