Given that
$x^2 y-x^3 \frac{d y}{d x}=y^4 \cos x$
On dividing by $y^4$, we get
$\begin{aligned} \frac{x^2}{y^3}-\frac{x^3}{y^4} \frac{d y}{d x} & =\cos x \\ \Rightarrow \quad \frac{x^3}{y^4} \frac{d y}{d x} & =\frac{x^2}{y^3}-\cos x\end{aligned}$
$\begin{array}{cc}\Rightarrow & \frac{1}{y^4} \frac{d y}{d x}-\frac{1}{x y^3}=-\frac{1}{x^3} \cos x \\ \text { Let } & -\frac{1}{y^3}=t\end{array}$
$\begin{array}{llrl}
\Rightarrow & \frac{1}{y^4} \frac{d y}{d x} & =\frac{1}{3} \cdot \frac{d t}{d x} \\
& \Rightarrow & \frac{1}{3} \cdot \frac{d t}{d x}+\frac{1 t}{x} & =\frac{1}{x^3} \cos x \\
& \Rightarrow & \frac{d t}{d x}+\frac{3}{x} t & =\frac{3}{x^3} \cos x
\end{array}$
This is a linear differential equation in $t$, on comparing with $\frac{d t}{d x}+P t=Q$, we get
$\begin{gathered}
P=\frac{3}{x}, Q=\frac{3}{x^3} \cos x \\
\therefore \quad \text { I.F. }=e^{\int P d x}=e^{\int \frac{3}{x} d x}=e^{\log x^3}=x^3
\end{gathered}$
$\therefore$ Complete solution is
$\begin{aligned}
& t x^3 & =3 \int \frac{x^3}{x^3} \cos x d x+c_1 \\
\Rightarrow & t x^3 & =3 \sin x+c_1 \\
\Rightarrow \quad & -\frac{1}{y^3} x^3 & =3 \sin x+c_1 \\
\Rightarrow \quad & y^{-3} x^3 & =-3 \sin x+c
\end{aligned}$