Given the following equation,
$\dot x(t) = e^{x(t)}$
The Laplace transform of $\dot x $ is $ sX(s) - x(0) $.
a) What is the Laplace transform (natural transform as well) of $e^{x(t)}$ ?
b) or Laplace transform of $\frac{1}{x(t)}$ (I think that it is $\frac{1}{X(s)})$
Although the Laplace transform of $\dot{x}$ can be computed, the transform of a generally nonlinear function of $x$ cannot be computed unless we already know $x(t)$. This is why the Laplace transform is only used to solve linear differential equations (ones of the form $a_0 x + a_1 \dot{x} + \ldots + a_n x^{(n)} = f(t)$, where $f(t)$ is some given function of $t$ which has a Laplace transform.
Statement b) is not correct - it would be
$$ \mathcal{L}\left\{\frac{1}{x}\right\}(s) = \int_0^\infty \frac{e^{-s\tau}}{x(\tau)}\mathrm{d}\tau, $$
which we cannot compute unless we know $x(t)$. Even so, this Laplace transform may not exist, even if the Laplace transform of $x$ does. The Laplace transform cannot be used on nonlinear mappings of $x$.
I have tried to find exceptions to that general rule, but I have not been able to find any. I think we can only compute the Laplace transforms of $x$, its derivatives, integrals, functions $t^n x$ and $x/t^n$ as well as convolutions of $x$ with known functions and $x$ itself.