How do I apply integrating factor to solve this differential equation? $$x \frac{dy}{dx}  +  y  =  -2x^6y^4$$
I tried to find the general solution by dividing both sides by $x$  or $x^6$ but no solution I could get.. Do I even solve it with integrating factor?

Contrary to other answers, you CAN find an integrating factor and manipulate your ODE via substitutions.
We have the ODE :
$$xy' + y = -2x^6y^4$$
Dividing both sides by $-\frac{1}{3}xy^4$, we yield :
$$-\frac{3y'}{y^4} - \frac{3}{xy^3} = 6x^5$$
Let $v(x) = \frac{1}{y^3(x)}$ and then this gives $v'(x) = -\frac{3y'(x)}{y^4(x)} $ and then the differential equation becomes :
$$v'(x) - \frac{3v(x)}{x} = 6x^5$$
Let $μ(x) = e^{\int -\frac{3}{x}\rm d x} = \frac{1}{x^3}$ and then multiply both sides by $μ(x)$ :
$$\frac{v'(x)}{x^3}-\frac{3v(x)}{x^4}=6x^2$$
By substituting $-\frac{3}{x^4}=\big(\frac{1}{x^3}\big)'$ we have :
$$\frac{v'(x)}{x^3}-\bigg(\frac{1}{x^3}\bigg)'v(x)=6x^2$$
Now, we shall apply the reverse product rule : $f\frac{\rm d g}{\rm d x}+g\frac{\rm d f}{\rm dx} = \frac{\rm{d}}{\rm d x}(f\;g)$ :
$$\int \frac{\mathrm{d}}{\mathrm{d}x}\bigg(\frac{v(x)}{x^3}\bigg)\mathrm{d}x=\int6x^2\mathrm{d}x \implies v(x) = x^3(2x^3+c_1)$$
Now, you can substitute $v(x) = \frac{1}{y^3(x)}$ and solve for $y(x)$ to yield the final result.
