Step 1: From the substitution $u=x^3$, $v=y^2$, the original equation $3x^5dx-y(y^2-x^3)dy=0$ reduces to the homogeneous equation $2u\,du+(u-v)\,dv=0$, i.e. $\dfrac{dv}{du}=\dfrac{2u}{v-u}$.
Step 2: Since this is homogeneous of degree one, put $v=tu$, so $dv=t\,du+u\,dt$. Then \[t\,du+u\,dt=\frac{2u}{tu-u}\,du=\frac{2}{t-1}\,du\]
Step 3: Isolate $dt$: \[u\,dt=\left(\frac{2}{t-1}-t\right)du=\frac{2-t(t-1)}{t-1}\,du=\frac{-(t-2)(t+1)}{t-1}\,du\] \[\frac{du}{u}=\frac{-(t-1)}{(t-2)(t+1)}\,dt\]
Step 4: Split into partial fractions: $\dfrac{t-1}{(t-2)(t+1)}=\dfrac{1/3}{t-2}+\dfrac{2/3}{t+1}$. Integrating, \[\ln u=-\frac13\ln(t-2)-\frac23\ln(t+1)+C\] \[u^3(t-2)(t+1)^2=K\]
Step 5: Since $t=v/u$, this gives the implicit solution \[(v-2u)(v+u)^2=K\]
Step 6: Differentiate implicitly with respect to $u$, treating $F(u,v)=(v-2u)(v+u)^2-K$: \[F_u=-6u(v+u), \qquad F_v=3(v+u)(v-u)\] \[\frac{dv}{du}=-\frac{F_u}{F_v}=\frac{6u(v+u)}{3(v+u)(v-u)}=\frac{2u}{v-u}=\frac{-2u}{u-v}\]
Step 7: This matches the given form $\dfrac{dv}{du}=\dfrac{\lambda u}{u-v}$, confirming \[\boxed{\lambda=-2}\]