Differential equation!

$(2xydx+x^2dy) + x^2ydx + (\frac {y^3}{3}dx + y^2dy)=0$

Putting $x^2y=t$ and $\frac {y^3}{3} = u$

$(dt+tdx)+(udx+du)=0$

Presence of $f(x) + f'(x)$ highly suggests to multiply both sides by $e^x$

$(e^xdt + t\cdot e^xdx)+ (u\cdot e^xdx+e^xdu)=0$

$\therefore, d(e^xt) + d(e^xu)=0$

$e^x(t+u)+c=0$ , where $c$ is the constant is integration.

Put in the values of $t$ and $u$ , use $y(1)=1$ to find $c$ , you should get $c=\frac {4e}{3}$ , and the value of $y^3(0)$ should come out to be $\large 4e$ .