$$2xy(1+y^2)dx - (1+x^2+x^2y^2)dy = 0$$
I haven't had my first test yet, so it's most likely that this equation is separable, homogeneous, exact, linear or Bernoulli. It would definitely be exact if there wasn't a 1+ on the dy side. But maybe there's something I am not seeing yet?
On
This equation becomes "nicer" if we start using $x=\sqrt z$ which makes $$2xy(1+y^2)\frac{dx}{dy} - (1+x^2+x^2y^2) = 0$$ $$y(1+y^2)\frac{dz}{dy}-(1+y^2)z-1=0$$ So, the homogeneous part is simple $$y(1+y^2)\frac{dz}{dy}-(1+y^2)z=0\implies y\frac{dz}{dy}-z=0\implies z=y+C$$ and, using the method of variation of parameters leads to $$C=-1-y\tan^{-1}(y)+K$$ making finally $$x=\pm \sqrt{Ky-1-y\tan^{-1}(y)}$$
Hint: $$\dfrac{\partial M}{\partial y}=2x+6xy^2$$ $$\dfrac{\partial N}{\partial x}=-2x-2xy^2$$ then the equation is not exact. For integral factor we see $$p(y)=\dfrac{M_y-N_x}{-M}=-2\dfrac{1+2y^2}{y(1+y^2)}$$ We have integral factor $$I=e^{\int p(y)dy}=\dfrac{1}{y^2(1+y^2)}$$ then the equation $$\dfrac{2x}{y}dx-\dfrac{1+x^2+x^2y^2}{y^2(1+y^2)}dy=0$$ is exact.