The general solution for this differential equation? Find the general solution of this differential equation:
$$ \frac{dy}{dx} = \frac{3x^5 y^3}{4} $$
Here's what I've done so far:
$ dy=\frac{3x^5 y^3 dx}{4} $
$ 4dy = 3x^5 y^3 dx $
$ \frac{4dy}{y^3} = 3x^5 dx $
Integrating both sides:
$ \int \frac{4dy}{y^3} = \int 3x^5 dx $
$ \int \frac{4dy}{y^3} = \frac{3x^6}{6} + C $
$ 4\int \frac{dy}{y^3} = \frac{x^6}{2} + C $
To solve the left side, let $ u=y^3 $, therefore $ du=3y^2 dy $, so:
$ \frac{4}{3} \int \frac{3y^2 dy}{y^5} = \frac{x^6}{2} + C $
$ \frac{4}{3} \int \frac{du}{y^5} = \frac{x^6}{2} + C $
$ \frac{4}{3} \int \frac{du}{y^3 y^2} = \frac{x^6}{2} + C $
$ \frac{4}{3} \int \frac{du}{u * y^2} = \frac{x^6}{2} + C $
I'm not sure what to do next. How do I get rid of that $ y^2 $? Or did I do something wrong?

Your equation is separable and you can rewrite it as $$\frac{dy}{y^3}=\frac{3}{4}x^5 dx$$ So, integrating both sides gives $$-\frac{1}{2 y^2}=\frac{x^6}{8}+C_1$$ which, after simplifications, write $$y^2=-\frac{4}{x^6+C_2}$$ that is to say $$y=\pm\frac{2}{\sqrt{C_3 -x^6}}$$
