$y \mid x$ let $x=ay$ then $4a^2=4ay+4(y^4+2y^2)$ thus $(2a-y)^2=y^2(4y^2+9)=\frac{(8y+9)^2-81}{16}$. thus $$(4(2a-y))^2=(8y^2+9)^2-81$$ So $(8y^2+9+8a-4y)(8y^2+9-8a+4y)=81$. -We cannot have $8y^2+9+8a-4y=-1$ or $8y^2+9-8a+4y=-1$. -If $8y^2+9+8a-4y=1$ then $8y^2+9-8a+4y=81$ and $16y^2+18=82$ thus $y^2=4$ and if $8y^2+9-8a+4y=1$ then with the same way $y^2=4$ and thus $y-2a=10$ or $y-2a=-10$ $y=2$ leads to $a=-4$ or $a=6$ and thus $x=-8$ or $x=12$. $y=-2$ leads to $a=-6$ or $a=4$ and thus $x=12$ or $x=-8$ Thus $8y^2+9+8a-4y=8y^2+9-8a+4y=-9$ or $8y^2+9+8a-4y=8y^2+9-8a+4y=9$ : -The first one leads to $4(2y^2+2a-y)=-18$ which means that $4$ divides $18$ but it is not possible, -the second one leads to $8y^2+8a-4y=8y^2-8a+4y=0$ thus $2a=y$ and $y=0$. Hence the only solution is $(0,0),(-8,2),(12,2),(-8,-2),(12,-2)$.

$x^2-x(y^2)-(y^6+2y^4)=0,\:\Delta=y^4(4y^2+9)=l^2$ $\,\Rightarrow\,4y^2+9=m^2\iff(m-2y)(m+2y)=9$. $(x,y)=(0,0)$ is a solution. Otherwise $y\neq 0$ and $m>3$ or $m<-3$, $m$ odd. So $m\in\{-7,-5,5,7\}$. $m\in\{-7,7\}$ is impossible. $m\in\{-5,5\}\,\Rightarrow\, y\in\{-2,2\}$, which gives $(x,y)=(12,2),(-8,2),(12,-2),(-8,-2)$.

$x^2-xy^2-(y^6+2y^4)=0.$The discrimimant of this equation must be p.square. $D=4y^6+9y^4=y^4(4y^2+9)$ $y^4$ is p.s.So $4y^2+9=m^2$ $(m-2y)(m+2y)=1*9=9*1=3*3$

already solved see here: http://www.artofproblemsolving.com/community/c4h1228945p6199107

The problem is poorly stated. You must write: Find integral solutions of the equation... I spent hours solve it in reals!

$x=yk$ $k= \sqrt{yk+y^4+2y^2}$ $k=yn$ $x=y^2n$ $n^2=n+y^2+2$ $(n-y)(n+y)=n+2$ Let $y,n \in Z^+$ if $n-y\in Z^-$ its imposible because rightside is positive $y>2$ $\Rightarrow$ $n+y>n+2$ $y=1,2$ easy check and we could have $y=2,n=3,x=12$ Lets $x,y \in Z^-$ $n=-k,y=-c$$c,k \in N$ $(-k-c)(c-k)=2-c$$\Rightarrow$ $(c-k)(c+k)=c-2$,$c>k$,$c+k>c-k$but $c=2$ so $k=2,x=-8$ lets $n\in N,y=-k$ $n>k,n+k>n+2 $ if $k>2$ so $k=1,2 $ check this variant its like $n,y \in N $ so $x=12,y=-2$ $y\in N ,n=-k$ its like $n,y \in Z^-$ so answer is $n=-2,y=2,x=-8$ and dont forget they can equal $0$ so we have second $x,y=(0,0)$