The solution of the differential equation $x^3 \frac{d y}{d x}+4 x^2 \tan y=e^x \sec y$ satisfying $y(1)=0$, is

$\sin y=e^x(x-1) x^{-4}$

We have,

$x^3 \frac{d y}{d x}+4 x^2 \tan y=e^x \sec y$

$\Rightarrow x^3 \cos y \frac{d y}{d x}+4 x^2 \sin y=e^x $

$\Rightarrow x^4 \cos y d y+4 x^3 \sin y d x=x e^x d x$

$\Rightarrow d\left(x^4 \sin y\right)=x e^x d x$

On integrating, we get

$x^4 \sin y=(x-1) e^x+C$            .......(i)

It is given that $y=0$ where $x=1$.

Putting $x=1$ and $y=0$ in (i), we get $C=0$

Putting $C=0$ in (i), we get

$x^4 \sin y=(x-1) e^x \Rightarrow \sin y=e^x(x-1) x^{-4}$