Let $y=y\left(x\right)$ be the solution of the differential equation $(3y^2-5x^2)ydx+2x(x^2-y^2)dy=0$ such that  $y\left(1\right)=1$. then $|{\left(y\left(2\right)\right)}^3-12y\left(2\right)|$ is equal to : 

$$\begin{gathered} (3y^2-5x^2)y.dx+2x(x^2-y^2)dy=0 \\ \Rightarrow \frac{dy}{dx}=\frac{y(5x^2-3y^2)}{2x(x^2-y^2)} \\ Put, y=mx \\ \Rightarrow m+x.\frac{dm}{dx}=\frac{m(5-3m^2)}{2(1-m^2)} \\ x.\frac{dm}{dx}=\frac{(5-3m^2)m-2m(1-m^2)}{2(1-m^2)} \\ \Rightarrow \frac{dx}{x}=\frac{2(m^2-1)}{m(m^2-3)}dm \\ \Rightarrow \frac{dx}{x}=(\frac{2}{m}-\frac{\frac{4}{3}}{m}+\frac{\frac{4m}{3}}{m^2-3})dm \\ \Rightarrow \int \frac{dx}{x}=\int \frac{\left(\frac{2}{3}\right)}{m}+\int \left(\frac{2m}{m^2-3}\right)dm \\ \Rightarrow \ln \left|x\right|=\frac{2}{3}\ln \left|m\right|+\frac{2}{3}\ln |m^2-3|+C \\ or \ln \left|x\right|=\frac{2}{3}\ln \left|\frac{y}{x}\right|+\frac{2}{3}\ln |{\left(\frac{y}{x}\right)}^2-3|+C \\ put, (x=1,y=1); we get, c=-\frac{2}{3}\ln \left(2\right) \\ \Rightarrow \ln \left|x\right|=\frac{2}{3}\ln \left|\frac{y}{x}\right|+\frac{2}{3}\ln |{\left(\frac{y}{x}\right)}^2-3|-\frac{2}{3}\ln \left(2\right) \\ \Rightarrow \left(\frac{y}{x}\right)[{\left(\frac{y}{x}\right)}^2-3]=2.\left(x^{\frac{3}{2}}\right) \end{gathered}$$

Put $x=2$ to get y (2)

$$\begin{gathered} \Rightarrow y(y^2-12)=4\times 2\times 2\times 2\sqrt{2} \\ \Rightarrow y^3-12y=32\sqrt{2} \\ \Rightarrow |y^3\left(2\right)-12y\left(2\right)|=32\sqrt{2} \end{gathered}$$