My solution: $x=0\implies y=0$ and $y=-1$ and at $y=\pm{8},\pm{4},,\pm{2},1$ no integer solution $\implies y\neq \pm{8},\pm{4},,\pm{2},1$ Then $y^3=8x^6+2x^3y-y^2\implies \frac{y^3}{x}=8x^5+2x^2y-\frac{y^2}{x}\implies \frac{y^2}{x}(y+1)=8x^5+2x^2y\implies x|y^2\implies x|y$ Then $y^2+y=\frac{8x^6}{y}+2x^3\implies y|8x^6\implies y|x$ because $y\neq \pm{8},\pm{4},,\pm{2},1$ THEN $x|y$ and $y|x\implies x=y$ THen we have $x^3=8x^6+2x^4-x^2\implies x=0$ The solution is $(x;y)=(0;0)(0;-1)$

I don't if there is any wrong in the solution but $(1,2)$ is also a solution.

Uzbekistan wrote: $\frac{y^2}{x}(y+1)=8x^5+2x^2y\implies x|y^2\implies x|y$ You don't know that $x$ and $y+1$ are coprime, and also $x\mid y^2$ doesn't generally imply $x\mid y$.

silouan wrote: Uzbekistan wrote: $\frac{y^2}{x}(y+1)=8x^5+2x^2y\implies x|y^2\implies x|y$ You don't know that $x$ and $y+1$ are coprime, and also $x\mid y^2$ doesn't generally imply $x\mid y$. Yeah you are right. I am sorry for my mistake

littletush wrote: Find all pairs of integers $(x,y)$ such that \[ y^3=8x^6+2x^3y-y^2 \]. HINT: We have $3y=6x^2+x-1+O(x^{-1})$ as $x \to \infty$.

Who is known?

We have $8x^6=y(y^2+y-2x^3)$ Let $|y|$ is odd but not $1$ let max power of a prime factor dividing $y$ is $a$ and $x$ is $b$ note $a+min (a,3b)=6b$ so $a=3b$ and hence $y=x^3$... it has no solution. when $y$ is even ,using similar process $y=4x^3$.....no solution so checking $|x|=1,0$ ......$(1,2),(0,0)$ is solution. now when $|y|=1$ then checking $1$ and $-1$ we get $(0,-1)$ as another solution.

The equation $y^3=8x^6+2x^3y-y^2$ can be written as $(64x^3 + 8y)^2 = (8y)^2(8y+9)$, whence $8y+9=z^2$, with wlog $z\geq 1$. This leads to $(4x)^3 = \pm (z^2-9)(z \mp 1)$, and now a simple framing of the RHS between consecutive cubes leads to just checking values $z\in \{1,3,5,7,9\}$, with only possible solutions $(0,-1), (0,0), (1,2)$ for $(x,y)$.