Solve the Differential equation $x^3 \frac{dy}{dx}=y^3+y^2\sqrt{x^2+y^2}$ Solve the Differential equation $$x^3 \frac{dy}{dx}=y^3+y^2\sqrt{x^2+y^2}$$
i reduced the equation as
$$x^3\frac{dy}{dx}=y^3\left(1+\sqrt{1+\left(\frac{x}{y}\right)^2}\right)$$ $\implies$
$$\frac{x^3}{y^3}\frac{dy}{dx}=\left(1+\sqrt{1+\left(\frac{x}{y}\right)^2}\right) \tag{1}$$ Next put $$\frac{x}{y}=v$$ we get
$$ x=vy$$ then
$$ \frac{dx}{dy}=v+y\frac{dv}{dy}$$ Then $(1)$ becomes
$$\frac{v^3}{v+y\frac{dv}{dy}}=1+\sqrt{1+v^2}$$ Reciprocating we get
$$\frac{v+y\frac{dv}{dy}}{v^3}=\frac{1}{\sqrt{1+v^2}+1}$$  Rationalizing RHS we get
$$\frac{v+y\frac{dv}{dy}}{v^3}=\frac{\sqrt{1+v^2}-1}{v^2}$$ Rearranging we get
$$\frac{dv}{v \times \left(\sqrt{1+v^2}-2\right)}=\frac{dy}{y}$$ 
EDIT: i am posting here the clue given by paul using Substitution $v=\tan z$:
$$\int\frac{dv}{v \times \left(\sqrt{1+v^2}-2\right)}=\int\frac{\sec^2 z\: dz}{\tan z(\sec z-2)}=\int\frac{dz}{\sin z(1-2\cos z)}$$ So
$$\int\frac{dz}{\sin z(1-2\cos z)}=\int \frac{\sin z\: dz}{\sin^2 z(1-2\cos z)}$$
Put $\cos z=t$ and use partial fractions

The solution of the ODE is presented below on parametric form. The explicit form can be obtained but involves the roots of a cubic polynomial equation which would lead to a big formula.

