How to solve the following homogeneous differential equation: (3x - 4y) + (2x - y)y' = 0

The final result must be: |y-?| = c|y+3x|^5

(3x - 4y) + (2x - y) y' = 0
y' = (4y - 3x) / (2x - y) . . . . . . . . [*]

A differential y' = f(x,y) is homogeneous if f(tx,ty) = f(x,y)
f(x,y) = (4y - 3x) / (2x - y)
f(tx,ty) = (4ty - 3tx) / (2tx - ty) = (t(4y - 3x)) / (t(2x - y)) = (4y - 3x) / (2x - y) = f(x,y)

So we do indeed have a homogeneous differential equation.
In that case, use substitution: u = y/x
y = xu
y' = xu' + u

Substituting y with xu and y' with xu'+u in equation [*], we get
xu' + u = (4xu - 3x) / (2x - xu)
xu' = x(4u - 3) / (x(2 - u)) - u
xu' = (4u - 3) / (2 - u) - u(2-u)/(2-u)
xu' = (4u - 3 - 2u + u²) / (2 - u)
xu' = (u² + 2u - 3) / (2 - u)
x du/dx = (u - 1)(u + 3) / (2 - u)
(2 - u) / ((u-1)(u+3)) du = 1/x dx

Using partial fraction decomposition on left side, we get:
((1/4) / (u-1) - (5/4) / (u+3)) du = 1/x dx
(1/(u-1) - 5/(u+3)) du = 4/x dx

Integrate both sides:
∫ (1/(u-1) - 5/(u+3)) du = ∫ 4/x dx
ln(u-1) - 5 ln(u+3) = 4 ln(x) + ln(C)
ln((u-1)/(u+3)⁵) = ln(Cx⁴)
(u-1) / (u+3)⁵ = Cx⁴

Now we substitute back: u = y/x
(y/x - 1) / (y/x + 3)⁵ = Cx⁴
[(y-x)/x] / [(y+3x)/x]⁵ = Cx⁴
(y-x)/x * x⁵/(y+3x)⁵ = Cx⁴
x⁴(y-x)/(y+3x)⁵ = Cx⁴
(y-x)/(y+3x)⁵ = C
(y-x) = C(y+3x)⁵