<s>[INST] Solving homogeneous differential equation $\frac{dy}{dx}=\frac{x^2+8y^2}{3xy}$. 
Solve the differential equation. Use the fact that the given equation is homogeneous
  $$
\frac{dy}{dx}=\frac{x^2+8y^2}{3xy}
$$

First I multiply the right side by
$$
\frac{\frac{1}{x^2}}{\frac{1}{x^2}}
$$
Then
$$\frac{dy}{dx}=\frac{1+\frac{8y^2}{x^2}}{\frac{3y}{x}}
$$
$$
Let: v=\frac{y}{x}
$$ $$
\frac{dy}{dx}=v+x\frac{dv}{dx}
$$
Then substitute
$$
\frac{dy}{dx}=\frac{1+v^2}{3v}$$
$$
\frac{1+v^2}{3v}=v+x\frac{dv}{dx}$$
How do I find the solution from here?
 [/INST] Hint:
First a slight miscalculation (you missed the $8$) on your part, it should be:
$$
\frac{1+8v^2}{3v}=v+x\frac{\mathrm{d}v}{\mathrm{d}x}$$
After this just separate the variables and integrate:
$$\int \frac{\mathrm{d}x}{x}=\int \frac{3v\ \mathrm{d}v}{(1+5v^2)}$$
Can you proceed?
Update:
To help you remove your confusion I have written down the remaining steps:
$$\frac{10}{3}\ln(xc)=\ln\left(1+5\left(\frac{y}{x}\right)^2 \right)$$
$$\ln\left((xc)^{\frac{10}{3}}\right)=\ln\left(1+5\left(\frac{y}{x}\right)^2 \right)$$
$$\frac 15 cx^{\frac{10}{3}}-\frac 15=\frac{y^2}{x^2}$$
$$\frac 15 cx^{\frac{10}{3}} \cdot x^2-\frac {x^2}{5}=y^2$$
$$\implies y=\pm \sqrt{\frac{cx^{\frac{16}{3}}}{5}-\frac {x^2}{5}}$$
$$\implies y=\pm \sqrt{c_1x^{\frac{16}{3}}-\frac {x^2}{5}}$$
</s>