Linear differential equation - inseparable? I need help with solving following linear differential equation:
$$y^\prime + \dfrac{3y}{x} = \dfrac{3}{x^4}$$

I have set the right side equal to $0$ and started solving:
$$y^\prime + \dfrac{3y}{x} = 0$$
$$y^\prime = - \dfrac{3y}{x}$$
$$\dfrac{dy}{dx}\times \dfrac{1}{y} = - \dfrac{3}{x}$$
$$\int\dfrac{1}{y} dy = -3\int \dfrac{1}{x} dx$$
$$\ln(y)=-3\times \ln(x) + c$$
And finally solution with left side (lets say this $c$ is: $e^c$)
$$y = c\times(-3x)$$

Using some formulas
$$y = c\times f(x)$$
$$y = c(x)\times f(x)$$
$$y^\prime = c^\prime(x)\times f(x)+c(x)\times f^\prime (x)$$
$$y = c(x)\times (-3x)$$
$$y^\prime = c^\prime(x)\times (-3x)+c(x)\times (-3)$$
And I replaced $y$ and $y^\prime $ in original equation and get
$$c^\prime(x)\times (-3x)+c(x)\times (-3)+\dfrac{3\times (c(x)\times(-3x))}{x} = \dfrac{3}{x^4}$$
And after adjustment:
$$\dfrac{c^\prime(x)\times (-3x)+c(x)\times (-3)+3\times (c(x)\times(-3x))}{x} = \dfrac{3}{x^5}$$
And what are next steps? I know I need to get $c(x)$ but how will I obtain it and what will be the final result?

By multiplying both sides by the integrating factor $x^3$, we get 
$$D(x^3y(x))=x^3y'(x) + 3 x^2 y(x) = \frac{3}{x}.$$
Hence
$$x^3y(x)=\int \frac{3}{x}\, dx=3\ln|x|+C,$$
that is
$$y(x)=\frac{3\ln|x|+C}{x^3}.$$
