Find all $x$ and $y$ such that $10xy^2=-x^3+4x$ and $13x^2y=-2y^3+4y$ 
Find all $x$ and $y$ such that $10xy^2=-x^3+4x$ and $13x^2y=-2y^3+4y$.

Well $(0,0)$ is a solution. And we can simplify as:
$10y^2=-x^2+4$ and $13x^2=-2y^2+4$
this is equal to:
$10y^2+x^2-4=0$ and $13x^2+2y^2-4=0$
and therefore
$10y^2+x^2-4=13x^2+2y^2-4$
which gives
$8y^2-12x^2=0$
or 
$2y^2-3x^2=0$
But I feel like I'm doing something wrong.

$\displaystyle10xy^2=-x^3+4x\iff x(10y^2+x^2-4)=0$
If $\displaystyle x=0,0=-2y^3+4y\iff2y(y^2-2)=0\implies y=0$ or $y^2=2$
Else $10y^2+x^2-4=0\  \ \ \ (1)$
Similarly start from $\displaystyle13x^2y=-2y^3+4y\iff y(13x^2+2y^2-4)=0$
Finally solve for $x^2,y^2$ from $(1)$ and $13x^2+2y^2-4=0$
