Differential equation of non standard form. Solve the differential equation:
$$\frac{dy}{dx}=\frac{x^{2}-y^{2}}{x^{2}(y^{2}+1)}$$

I tried to convert it into an exact differential but I failed to do so. I also tried to bring the equation into standard forms but again failed. Please help me. 
Here's the exact question:

 A: HINT for the last two options : Sketch $y(x)$ in $x>1$
One have to consider separately the cases : first $y(1)<-1$ , second $-1<y(1)<1$ , third $1<y(1)$. Consider the sign of $\frac{dy}{dx}$ which is the same as $|x|-|y|$ and the change of sign which indicate a maximum or minimum of $y(x)$. 
For $x$ tending to infinity, the equation tends to  $\frac{dy}{dx}\sim \frac{1}{y^2+1}$ leading to $\frac{1}{3}y^3\sim x$.

Of course, one must not give a graphical representation as a final answer. It is only a preliminary approach which makes understandable the behaviour of the function. Then, it is easier to find what will be the answer and how to prove it.
A: Option (A): Note that if $x>1$, then
$$ y'=\frac{x^2-y^2}{x^2(y^2+1)}=\frac{1}{y^2+1}- \frac{y^2}{x^2(y^2+1)}\ge\frac{1}{y^2+1}-\frac{y^2}{y^2+1}=-\frac{y^2-1}{y^2+1}. $$
Case 1: $|y|< 1$. Then
$$ \left[1+\frac{1}{y-1}-\frac{1}{y+1}\right]dy \le -x $$
and hence
$$ y+\ln\bigg|\frac{y-1}{y+1}\bigg|\le -(x-1)+y(1)+\ln\bigg|\frac{y(1)-1}{y(1)+1}\bigg|. $$
So
$$ e^y\bigg|\frac{y-1}{y+1}\bigg|\le ke^{-(x-1)}. $$
It is easy to see that 
$$ \lim_{x\to\infty} y(x)=1. $$
