A certain ellipse is centered at (0, 0) from which the major axis always equal...

$fdx-1gdy=0$
Where $f=(1-xy{)}^{(-2)}and9=(y^2+x^2(1-xy{)}^{(-2)})$
Satisfy $\frac{df}{dy}=\frac{dg}{dx}$
Suppose the solution is $F(x,y)=c$
Then, calculating the differential of both sides
$\frac{dF}{dx}dx+\frac{dF}{dy}dy=0$
So we see $f=\frac{dF}{dx}$
$g=\frac{dF}{dy}$
We can integrate these equations partially to find f. let's go for the y integral
$\frac{dF}{dy}=y^2+x^2/(1-xy{)}^2$
$F=y^3/3-x^2/(1-xy)(-1/x)+1x(x)$
$y^3/3+x/(1-xy)+x(x)$
where x(x) is an arbitrary function dx only.
Now substitute that into the equation
for $dF/dx$ to determine x(x)
$\frac{dF}{dx}=1/(1-xy{)}^2+dx/dx$
Comparing what with the coefficient of dx we get that $dx/dx=0sox(x)$ is just a constant which we can take to be 0.
So the solution is F=c or
$y^3/3+x/(1-xy)=c$
$\Rightarrow y^3+3x/(1-xy)=3c$
$\Rightarrow (1-xy)y^3+3x=3c(1-xy)$
$\Rightarrow y^3-x{y}^4+3x-3c\times 4=-3c$
$\Rightarrow x{y}^4-y^3+3cxy-3x=3c$