If  $x^{2}ydx-\left(x^3+y^3\right)dy=0$, $y\left(0\right)=1$  and  $y^3\log y=k{x}^3$ then k = 

$\begin{array}{l}\Rightarrow \frac{dy}{dx}=\frac{x^{2}y}{x^3+y^3}=\frac{y/x}{1+(y/x{)}^3}\\ V+x\frac{dv}{dx}=\frac{v}{1+v^3}(\text{ put }y=vx)\\ \Rightarrow x\frac{dv}{dx}=\frac{-v^4}{1+v^3}\\ \int \frac{1+v^3}{v^4}dv=-\int \frac{dx}{x}\\ \Rightarrow \int \left(v^{-4}+\frac{1}{v}\right)dv=-\int \frac{1}{x}dx\\ \Rightarrow \frac{v^{-3}}{-3}+\ln v=-\ln x+c\end{array}$

$\begin{array}{l}\Rightarrow \ln \left|x\left(\frac{y}{x}\right)\right|-\frac{1}{3v^3}=c\\ \ln y-\frac{x^3}{3y^3}=c\\ y\left(0\right)=1\Rightarrow c=0 \end{array}$

$\therefore \log y=\frac{x^3}{3y^3}$