$$y'=\frac{y}{x-2x^2y} \Rightarrow \frac{dy}{dx}=\frac{y}{x-2x^2y} \Rightarrow (x-2x^2y)dy=ydx \\ \Rightarrow ydx-(x-2x^2y)dy=0 \\ \Rightarrow M=y, N=-(x-2x^2y) \Rightarrow \frac{\partial{M}}{\partial{y}}=1, \frac{\partial{N}}{\partial{x}}=-1+4xy $$
Also $$\frac{\partial{M}}{\partial{y}} \neq \frac{\partial{N}}{\partial{x}}$$
$$\frac{M_y-N_x}{N}=\frac{1-(4xy -1)}{-(x-2x^2y) }=\frac{2-4xy}{-x+2x^2y} =\frac{2(1-2xy)}{-x(1-2xy)}=-\frac{2}{x}$$
Also der integrierende Faktor ist der folgende:
$$\mu (x,y)=e^{\int -\frac{2}{x} dx}=e^{-2\ln x}=e^{\ln x^{-2}}=x^{-2}=\frac{1}{x^2}$$