Differential Equation By Method of Integrating factors The differential equation is,

$$\frac{dy}{dx}=x^2-2x^2y$$

I have calculated $μ(x)$ as $μ(x)=e^{2x^3/3}$

thus 

$$y=e^{-2x^3/3}\int x^2e^{2x^3/3}\,dx$$

After this I have been having a bit of trouble solving for $y$. I have tried integration by parts but it hasn't helped, probably due to a mistake on my part.
If you are asking how to solve the integral you have, you can use u-substitution.

$u = \frac{2x^3}{3}$

$du = 2x^2$

So

$\int x^2e^{\frac{2x^3}{3}} = \frac{1}{2} \int e^u du = \frac{1}{2} e^{\frac{2x^3}{3}} + C$
If you're going to integrate by factors, couldn't you just write the question as:

$dy/dx = x^2(1-2y)$

This makes it much easier, because you now can integrate x and y separately, as it's own thing, etc.

$(1-2y)^{-1}\ dy=x^2\ dx$

Integrate both sides to get:

$-\frac12 \ln(1-2y)=x^3/3 + C_1$

$\ln(1-2y) = -\frac{2x^3}3 + C_2$

$1-2y=Ae^{-2x^3/3}$

$y=\frac12(1-Ae^{-2x^3/3})$