Differential equation xydx-(x+2y)^2dy=0

Given the equation:
$$- x^{2} \frac{d}{d x} y{\left(x \right)} - 4 x y{\left(x \right)} \frac{d}{d x} y{\left(x \right)} + x y{\left(x \right)} - 4 y^{2}{\left(x \right)} \frac{d}{d x} y{\left(x \right)} = 0$$
Do replacement
$$u{\left(x \right)} = \frac{y{\left(x \right)}}{x}$$
and because
$$y{\left(x \right)} = x u{\left(x \right)}$$
then
$$\frac{d}{d x} y{\left(x \right)} = x \frac{d}{d x} u{\left(x \right)} + u{\left(x \right)}$$
substitute
$$- 4 x^{2} u^{2}{\left(x \right)} \frac{d}{d x} x u{\left(x \right)} - 4 x^{2} u{\left(x \right)} \frac{d}{d x} x u{\left(x \right)} + x^{2} u{\left(x \right)} - x^{2} \frac{d}{d x} x u{\left(x \right)} = 0$$
or
$$- 4 x^{3} u^{2}{\left(x \right)} \frac{d}{d x} u{\left(x \right)} - 4 x^{3} u{\left(x \right)} \frac{d}{d x} u{\left(x \right)} - x^{3} \frac{d}{d x} u{\left(x \right)} - 4 x^{2} u^{3}{\left(x \right)} - 4 x^{2} u^{2}{\left(x \right)} = 0$$
This differential equation has the form:

f1(x)*g1(u)*u' = f2(x)*g2(u),

where
$$\operatorname{f_{1}}{\left(x \right)} = 1$$
$$\operatorname{g_{1}}{\left(u \right)} = 1$$
$$\operatorname{f_{2}}{\left(x \right)} = - \frac{1}{x}$$
$$\operatorname{g_{2}}{\left(u \right)} = \frac{4 \left(u{\left(x \right)} + 1\right) u^{2}{\left(x \right)}}{4 u^{2}{\left(x \right)} + 4 u{\left(x \right)} + 1}$$
We give the equation to the form:

g1(u)/g2(u)*u'= f2(x)/f1(x).

Divide both parts of the equation by g2(u)
$$\frac{4 \left(u{\left(x \right)} + 1\right) u^{2}{\left(x \right)}}{4 u^{2}{\left(x \right)} + 4 u{\left(x \right)} + 1}$$
we get
$$\frac{\left(4 u^{2}{\left(x \right)} + 4 u{\left(x \right)} + 1\right) \frac{d}{d x} u{\left(x \right)}}{4 \left(u{\left(x \right)} + 1\right) u^{2}{\left(x \right)}} = - \frac{1}{x}$$
We separated the variables x and u.

Now, multiply the both equation sides by dx,
then the equation will be the
$$\frac{dx \left(4 u^{2}{\left(x \right)} + 4 u{\left(x \right)} + 1\right) \frac{d}{d x} u{\left(x \right)}}{4 \left(u{\left(x \right)} + 1\right) u^{2}{\left(x \right)}} = - \frac{dx}{x}$$
or
$$\frac{du \left(4 u^{2}{\left(x \right)} + 4 u{\left(x \right)} + 1\right)}{4 \left(u{\left(x \right)} + 1\right) u^{2}{\left(x \right)}} = - \frac{dx}{x}$$

Take the integrals from the both equation sides:
- the integral of the left side by u,
- the integral of the right side by x.
$$\int \frac{4 u^{2} + 4 u + 1}{4 u^{2} \left(u + 1\right)}\, du = \int \left(- \frac{1}{x}\right)\, dx$$
Detailed solution of the integral with u
Detailed solution of the integral with x
Take this integrals
$$\frac{3 \log{\left(u \right)}}{4} + \frac{\log{\left(u + 1 \right)}}{4} - \frac{1}{4 u} = Const - \log{\left(x \right)}$$
Detailed solution of the equation
We get the simple equation with the unknown variable u.
(Const - it is a constant)

Solution is:
$$\operatorname{u_{1}} = \log{\left(x \right)} + \frac{\log{\left(u{\left(x \right)} + 1 \right)}}{4} + \frac{3 \log{\left(u{\left(x \right)} \right)}}{4} - \frac{1}{4 u{\left(x \right)}} = C_{1}$$
do backward replacement
$$y{\left(x \right)} = x u{\left(x \right)}$$
$$y1 = y(x) = C_{1} x$$