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

$x^3=3 y^3\left(1-e^{-x y}\right)$

The given differential equation is

$(y d x+x d y) y^4=x^2 e^{x y}(y d x-x d y)$

$\Rightarrow e^{-x y} d(x y)=\left(\frac{x}{y}\right)^2 d\left(\frac{x}{y}\right)$

Integrating, we obtain

$-e^{-x y}=\frac{1}{3}\left(\frac{x}{y}\right)^3+C$

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

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

Putting $C=-1$ in (i), we obtain

$-3 y^3 e^{-x y}=\left(x^3-3 y^3\right)$ or, $x^3=3 y^3\left(1-e^{-x y}\right)$

as the solution.