Keep the denominator sign in view

That is only valid when $x+3>0$. If $x+3<0$ the inequality reverses, so you would need two separate cases.

$x<-3$ and $0<x<5$ cannot both hold, giving the empty set. The solution is the union of the two intervals.

The original fraction is undefined there.

$\ln x$ only exists for positive $x$, so the interval $x<e^{-3}$ becomes $0<x<e^{-3}$.

$-3\to e^{-3}$, $0\to1$ and $5\to e^5$.