Follow the hints.
1. Let [tex]u=\dfrac yx[/tex], so that [tex]y=ux[/tex] and [tex]\mathrm dy=x,\mathrm du+u\,\mathrm dx[/tex]. Substituting into
[tex](xy+y^2)\,\mathrm dx-x^2\,\mathrm dy=0[/tex]
gives
[tex](ux^2+u^2x^2)\,\mathrm dx-x^2(x\,\mathrm du+u\,\mathrm dx)=0[/tex]
[tex]u^2x^2\,\mathrm dx-x^3\,\mathrm du=0[/tex]
and the remaining ODE is separable:
[tex]x^3\,\mathrm du=u^2x^2\,\mathrm dx\implies\dfrac{\mathrm du}{u^2}=\dfrac{\mathrm dx}x[/tex]
Integrate both sides to get
[tex]-\dfrac1u=\ln|x|+C[/tex]
[tex]-\dfrac xy=\ln|x|+C[/tex]
[tex]\boxed{y=\dfrac x{Cx-\ln|x|}}[/tex]
2. [tex]Let [tex]v=\dfrac xy[/tex], so that [tex]x=vy[/tex] and [tex]\mathrm dx=v\,\mathrm dy+y\,\mathrm dv[/tex]. Then
[tex]2x^2y\,\mathrm dx=(3x^3+y^3)\,\mathrm dy[/tex]
becomes
[tex]2v^2y^3(v\,\mathrm dy+y\,\mathrm dv)=(3v^3y^3+y^3)\,\mathrm dy[/tex]
[tex]2v^3y^3\,\mathrm dy+2v^2y^4\,\mathrm dv=(3v^3y^3+y^3)\,\mathrm dy[/tex]
[tex]2v^2y^4\,\mathrm dv=(v^3y^3+y^3)\,\mathrm dy[/tex]
which is separable as
[tex]\dfrac{2v^2}{v^3+1}\,\mathrm dv=\dfrac{\mathrm dy}y[/tex]
Integrating both sides gives
[tex]\dfrac23\ln|v^3+1|=\ln|y|+C[/tex]
[tex]\ln|v^3+1|=\dfrac32\ln|y|+C[/tex]
[tex]v^3+1=Cy^{3/2}[/tex]
[tex]v=\sqrt[3]{Cy^{3/2}-1}[/tex]
[tex]\dfrac xy=\sqrt[3]{Cy^{3/2}-1}[/tex]
[tex]\boxed{x=y\sqrt[3]{Cy^{3/2}-1}}[/tex]
