If $x^{2}y-x^3\frac{dy}{dx}=y^4\cos x$ then $x^3{y}^{-3}$ equal to

$x^{2}y-x^3\frac{dy}{dx}=y^4\cos x$
$\frac{x^2}{y^3}-\frac{x^3}{y^4}\frac{dy}{dx}=\cos x$
$\frac{1}{x{y}^3}-\frac{1}{y^4}\frac{dy}{dx}=\frac{\cos x}{x^3}$
$\frac{1}{y^4}\frac{dy}{dx}-\frac{1}{x{y}^3}=-\frac{1}{x^3}\cos x$
Let $t=\frac{-1}{y^3}$  
$\frac{1}{y^4}\frac{dy}{dx}=\frac{1}{3}\frac{dt}{dx}$
$\frac{1}{3}\frac{dt}{dx}+\frac{t}{x}=\frac{-1}{x^3}\cos x$
$\frac{dt}{dx}+\frac{3}{x}t=\frac{-3}{x^3}\cos x$
$I.F=X^3$
$t{x}^3=\int \frac{-3}{x^3}\cos x{x}^{3}dx+c$
$y^{-3}{x}^3=3\sin x+c$