Struggles solving : $\frac23xyy'=\sqrt{x^6-y^4}+y^2.$ 
Solve $\frac23xyy'=\sqrt{x^6-y^4}+y^2.$
(Hint: the equation can be boiled down to a homogeneous one.)

My attempt:
I didn't know how to obtain a homogeneous equation, so I'm going to show what I did instead.
We can write $\frac23xyy'=|x|^3\sqrt{1-\frac{y^4}{x^6}}+y^2.$ Let $z=\frac{y^2}{x^3}=y^2x^{-3}.$ Then
$$\begin{aligned}z'&=2yy'x^{-3}-3y^2x^{-4}\\ \implies x^4z'&=2xyy'-3y^2\\ \implies\frac23xyy'&=\frac{x^4z'+3y^2}3=\frac{x^4z'}3+y^2\\ \implies\frac{x^4z'}3+y^2&=|x|^3\sqrt{1-z^2}+y^2\\ \implies\frac{|x|z'}3&=\sqrt{1-z^2}\\ \implies\frac{dz}{\sqrt{1-z^2}}&=3\frac{dx}{|x|}\\ \implies\arcsin z&=3\ln|x|+C, C\in\Bbb R\\ \implies z&=\sin\ln(C|x|^3), C>0\\ \implies y^2&=x^3\sin\ln(C|x|^3)C>0.\end{aligned}$$
I'm not sure my answer is correct. I didn't get far plugging the result into the equation.
How do we solve the given ODE? Or should I better ask, how do we obtain a homogeneous ODE from it?

We can obtain the homogenious equation as follows.
$$\begin{align}
&\Longleftrightarrow \frac{x}{3}\cdot \frac{d(y^2)}{dx} = \sqrt{(x^3)^2-y(^2)^2}+y^2\\
&\Longleftrightarrow x^3\cdot \frac{d(y^2)}{3x^2dx} = \sqrt{(x^3)^2-y(^2)^2}+y^2\\
&\Longleftrightarrow x^3\cdot \frac{d(y^2)}{d(x^3)} = \sqrt{(x^3)^2-y(^2)^2}+y^2\\
&\Longleftrightarrow \frac{d(y^2)}{d(x^3)} = \frac{\sqrt{(x^3)^2-y(^2)^2}+y^2}{x^3} \tag{1}\\
\end{align}$$
Let's denote $(p,t) = (y^2,x^3)$ then
$$(1) \Longleftrightarrow  \frac{dp}{dt} = \frac{\sqrt{t^2-p^2}+p}{t} \tag{2}$$
$(2)$ is homogenious and you can solve it easily (for example, transform $(2)$ to an equation of $\frac{p}{t}$).
