Here, the differential equation
$$
\begin{aligned}
& \left(x-4 y^3\right) \frac{d y}{d x}-y=0,(y>0) \\
& \Rightarrow \quad\left(x-4 y^3\right) \frac{d y}{d x}=y \quad \Rightarrow \quad \frac{d x}{d y}=\frac{x-4 y^3}{y} \\
& \Rightarrow \quad \frac{d x}{d y}=\frac{x}{y}-4 y^2 \quad \Rightarrow \quad \frac{d x}{d y}-\frac{x}{y}=-4 y^2 \\
& \therefore \quad P=-\frac{1}{y} \cdot Q=-4 y^2
\end{aligned}
$$
Integrating factor,
$$
\text { I.F. }=e^{\int P d y}=e^{\int-\frac{1}{y} d y}=e^{-\log y}=e^{\log y-1}=\frac{1}{y}
$$
Now, the solution is given by
$$
\begin{aligned}
& x \cdot(\text { I.F. })=\int(\text { I.F. }) \cdot Q d y+C \\
& \frac{x}{y}=\int \frac{1}{y} \times\left(-4 y^2\right) d y+C \\
& \frac{x}{y}=\int-4 y d y+C \\
& \frac{x}{y}=-4 \frac{y^2}{2}+C
\end{aligned}
$$
$$
\begin{aligned}
& x=-2 y^3+C y \\
& x+2 y^3=C y
\end{aligned}
$$
$\therefore \quad$ This is the required solution.