Solve this following

$\left(x-x^{3}\right) d y=\left(y+y x^{2}-3 x^{4}\right) d x$

$\Rightarrow x d y-y d x=\left(y x^{2}-3 x^{4}\right) d x+x^{3} d y$

$\Rightarrow \frac{x d y-y d x}{x^{2}}=(y d x+x d y)-3 x^{2} d x$

$\Rightarrow \mathrm{d}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)=\mathrm{d}(\mathrm{xy})-\mathrm{d}\left(\mathrm{x}^{3}\right)$

Integrate

$\Rightarrow \frac{y}{x}=x y-x^{3}+c$

given $f(3)=3$

$\Rightarrow \frac{3}{3}=3 \times 3-3^{3}+\mathrm{c}$

$\Rightarrow \mathrm{c}=19$

$\therefore \frac{y}{x}=x y-x^{3}+19$

at $x=4, \frac{y}{4}=4 y-64+19$

$15 y=4 \times 45$

$\Rightarrow \mathrm{y}=12$