Asseev

1)
$$y'=\sqrt{\frac{x^2+x^2y^2}{y^2+x^2y^2}}$$
$$\frac{dy}{dx}=\sqrt{\frac{x^2(1+y^2)}{y^2(1+x^2)}}$$
$$\frac{dy}{dx}=\frac{x}{y}\sqrt{\frac{(1+y^2)}{(1+x^2)}}$$$$\frac{ydy}{\sqrt{1+y^2}}=\frac{xdx}{\sqrt{1+x^2}}$$
$$\int\frac{ydy}{\sqrt{1+y^2}}=\int\frac{xdx}{\sqrt{1+x^2}}$$
Ahora hacemos las sustituciones simples,
$$u=1+x^2$$$$\frac{du}{2}=x$$

$$t=1+y^2$$
$$\frac{dt}{2}=y$$

Con esto tenemos,
$$\frac{1}{2}\int\frac{dt}{\sqrt{t}}=\frac{1}{2}\int\frac{du}{\sqrt{u}}$$
$$\sqrt{1+y^2}+k=\sqrt{1+x^2}+q$$ $$\sqrt{1+y^2}=\sqrt{1+x^2}+c$$ 
$$1+y^2=1+x^2+2\sqrt{1+x^2}c+c^2$$
$$y=\sqrt{x^2+2\sqrt{1+x^2}c+c^2}$$
(con q,k,c constantes)
2) 
$$(3x^2\tan({y})-\left(\displaystyle\frac{2y^3}{x^3} \right))dx+(x^3\sec^2({y})+4y^3+\frac{3y^2}{x^2})dy=0$$
 
$$\displaystyle\frac{{\partial p}}{{\partial y}}=\sec^2({y})\cdot{}3x^2-\left( \displaystyle\frac{6y^2}{x^3}\right)$$
 
 
$$\displaystyle\frac{{\partial q}}{{\partial x}}=3x^2\sec^2({y})+\left( \displaystyle\frac{6xy}{x^4} \right)$$
 
                                   Exacta
 
 Integrando
 
 
$$g(x,y)=\displaystyle\int(3x^2\tan({y})-\left( \displaystyle\frac{2y^3}{x^3} \right))dx+h( y )$$
 
$$g(x,y)=3\tan({y})\left( \displaystyle\frac{x^3}{3}\right)+2y^3\left( \displaystyle\frac{x^-2}{-2} \right)$$
 
$$g(x,y)=\tan({y})x^3+\left( \displaystyle\frac{y^3}{x2} \right)+h( y )/\displaystyle\frac{{\partial }}{{\partial y}}$$
 $$Q(x,y)= \sec^2({y})+x^3+3y^2x^-2+h'( y )$$
 
$$x^3\sec^2({y})+4y^3+\left( \displaystyle\frac{3y^2}{x^2} \right)=\sec^2({y})x^3+3y^2x^-2+h'( y )$$
 
$$h'( y )= 4y^3$$   integrando
 
$$h( y )= y^4+C$$
 
$$\tan({y})x^3+\left( \displaystyle\frac{y^3}{x2} \right)+y^4+c=0$$
 
$$c= -x^3\tan({y})-\left( \displaystyle\frac{y^3}{x2} \right)-y^4$$