Solve a differential equation using variable separation I am trying to solve the following differential equation:

$y'=xy^{2/3}$ with the initial condition: $y(0) = 0$

Here is my progress so far:

We separate the fariables: $\int y'y^{-2/3}dy = \int x dx$

Therefore, $\int y'y^{-2/3}dy = yy^{-2/3} + \dfrac{3}{2}\int yy^{-5/3} = y^{1/3} +\dfrac{3}{2}\int y^{-2/3} = y^{1/3} - 3\dfrac{3}{2}y^{1/3} = y^{1/3} = -\dfrac{9}{2}y^{1/3}$

I know the answer to the problem, and I am supposed to find $3y^{1/3} = 3x^2+C$ which is obviously different from my result
It is $y'=\frac{dy}{dx}$

Therefore the equation is 

$y'=xy^{2/3}\Rightarrow \frac{dy}{dx} =xy^{2/3}$

Multiplying both sides by $dx$

$dy =xy^{2/3} \, dx$

Now you can go on with seperating the variables.
$$y'=xy^{2/3}\implies \frac{dy}{dx} =xy^{2/3}\implies \frac{dy}{y^{2/3}}=x\,dx $$ So $$\int \frac{dy}{y^{2/3}}=\int x\,dx \implies 3 \sqrt[3]{y}=\frac 12 x^2+C$$ Then $$\sqrt[3]{y}=\frac 16 x^2+C\implies y=\left(\frac 16 x^2+C\right)^3$$