About the diophantine equation $y^3=8x^6+2x^3y-y^2$ 
How I can solve the equation $y^3=8x^6+2x^3y-y^2$ in integers?

I made the substition $x^3=z$ and got the equation $8z^2+2yz-y^3-y^2=0$. So I decided to apply the general formula for quadratic equations and thus I got $$z=\frac{-2y\pm \sqrt{32y^3+36y^2}}{16}=\frac{-y\pm y\sqrt{8y+9}}{8}.$$
So in order to have $z\in \mathbb{Z}$, $8y+9$ must be a perfect square, let's say $8y+9=k^2$, with $k\in \mathbb{Z}$. From this we get $8y=(k+3)(k-3)$, but what could I do next? Also, is there another approach, like an appropriate factorization, to solve this equation? Thanks in advance.

What you have so far means that $k=2m+3$, so $8y+9 = 4m^2+12m+9\Rightarrow y =\dfrac{m(m+3)}{2}\in\mathbb{Z}, \forall m$. Then, $x^3 = z = \dfrac{m(m+3)}{16}(-1\pm (2m+3) ) = \dfrac{m(m+1)(m+3)}{8}$ or $x^3 = -\dfrac{m(m+2)(m+3)}{8}$. 
Now, the only options for the first case are $2x = m+1,m+2 $ and you can do the same for the second equation as well. That should determine what $m$ should be and therefore give you all the solutions.  
