Solving the ODE $y(1+\sqrt{x^2 y^4+1})dx+2xdy=0$ Question: 

Solve the ODE given below:
  $y(1+\sqrt{x^2 y^4+1})dx+2xdy=0$   


My try: 
The equation is not separable because a function of $x$ is added to a function of $y$.
($y+y\sqrt{x^2y^4+1}$)
Also, it is not linear with respect to $x$ or $y$, because it has the term$\sqrt{x^2y^4+1}$.
On the other hand, it's not a complete ODE because $\frac{d}{dy}(y(1+\sqrt{x^2 y^4+1})) \neq \frac{d}{dx}(2x)$
I also tried homogenous ODE's. But this ODE is not homogenous.  It doesn't seem to be a Clero DE either. 

Any idea?
Thanks in advance.

$$y(1+\sqrt{x^2 y^4+1})dx+2xdy=0$$
$$2xyy'=-y^2(1+\sqrt{x^2 y^4+1})$$
$u(x)=xy^2 \quad\to\quad u'=y^2+2xyy'=y^2-y^2(1+\sqrt{x^2 y^4+1})=-y^2\sqrt{x^2 y^4+1})$
$$xu'=-xy^2\sqrt{x^2 y^4+1})=-u\sqrt{u^2+1}$$
$$\frac{u'}{u\sqrt{u^2+1}}=-\frac{1}{x}$$
$$\int\frac{du}{u\sqrt{u^2+1}}=-\int\frac{dx}{x}$$
$$\ln|u|-\ln|1+\sqrt{u^2+1}|=-\ln|x|+\text{constant}$$
$$\frac{u}{1+\sqrt{u^2+1}}=\frac{c}{x}$$
$$u=\frac{2cx}{x^2-c^2}$$
$$y^2=\frac{u}{x}=\frac{2c}{x^2-c^2}\quad\to\quad y=\pm\sqrt{\frac{2c}{x^2-c^2}}$$
Bringing back into the ODE shows that it agrees.
