Reduction of Order ODE

For the second differential equation : $$xy^\prime(y y^{\prime\prime}-(y^\prime)^2)-y(y^\prime)^2=x^4y^3$$ Divide by $y'y^2$: $$x\dfrac {(y y^{\prime\prime}-(y^\prime)^2)}{y^2}-\dfrac {y'}{y}=\dfrac {x^4y}{y'}$$ $$x\left(\dfrac {y'}{y}\right)'-\dfrac {y'}{y}=\dfrac {x^4y}{y'}$$ $$x(\ln y)''-(\ln y)'=\dfrac {x^4y}{y'}$$

$$ \left ( \dfrac {(\ln y)'}{x}\right )'=\dfrac {x^2y}{y'}$$ $$ (\ln y)' \left ( \dfrac {(\ln y)'}{x}\right )'=x^2$$ $$ \dfrac {(\ln y)'}{x}d \left ( \dfrac {(\ln y)'}{x}\right )=x dx$$ Integrate both sides. $$ \left( \dfrac {(\ln y)'}{x}\right )^2=x^2+C$$ $$ \left( {(\ln y)'}\right )^2=x^4+Cx^2$$ $$ \ln y=\pm \int \sqrt {x^4+Cx^2} dx$$

If your first equation is really $$y^2y^{\prime\prime\prime}-3yy^\prime y^{\prime\prime}+2(y^\prime)^3+\frac{y}{x}(yy^\prime-(y^\prime)^2)=\frac{y^3}{x^2},$$ then it can be written as: $$\left(\dfrac {y'}{y}\right)''+\frac{1}{x}\left(\left(\dfrac {y'}{y}\right)-\left(\dfrac {y'}{y}\right)^2\right)=\dfrac {1}{x^2}.$$ By using $\frac{y'}{y}=w+\frac{1}{2}$ we get $$w''-\frac{1}{x}w^2=\frac{4-x}{4x^2}.$$ And I don't see much hope for this to be integrable. But if your equation is $$y^2y^{\prime\prime\prime}-3yy^\prime y^{\prime\prime}+2(y^\prime)^3+\frac{y}{x}(yy''-(y^\prime)^2)=\frac{y^3}{x^2},$$ the it can be written as $$\left(\dfrac {y'}{y}\right)''+\frac{1}{x}\left(\dfrac {y'}{y}\right)'=\dfrac {1}{x^2}.$$ Using $\frac{y'}{y}=v$ we get $$v''+\frac{1}{x}v'=\frac{1}{x^2},$$ and then $v'=z$, we get $$z'+\frac{1}{x}z=\frac{1}{x^2}.$$ This is a first order linear ODE and can be solved with a integrating factor.