Note that the expression for the joint density $f_{X,Y}(x,y)$ that you computed only holds for $x$ and $y$ satisfying $x \ge 0$ and $y \ge \max\{x, 4\}$; it is zero otherwise. Thus for $x \ge 0$ we have
$$f_X(x) = \int_{-\infty}^\infty f_{X,Y}(x,z) \, dz = \int_{\max\{x, 4\}}^\infty \frac{192}{z^5} \, dz.$$
Continuing the computation leads to
$$f_X(x) = 48 \max\{x, 4\}^{-4} = \begin{cases}48 / 4^4 & x \in [0, 4] \\ 48 / x^4 & x > 4\end{cases}$$
So,
$$\int_0^\infty f_X(x) \, dx = 4 \cdot \frac{48}{4^4} + \int_4^\infty \frac{48}{x^4} \, dx
= \frac{3}{4} + \frac{16}{4^3} = 1.$$
The cdf $F_X$ always exists for every random variable, which is not necessarily true of the density. So, yes, technically it is always possible to obtain the density from the cdf, if the density exists.
In general, there are many different ways to characterize a probability distribution, and some of them are more tractable than others depending on the situation. In certain situations, it may be easiest to characterize the density by, e.g., deriving the moment-generating function of the random variable. For example, if you find that the mgf corresponds to the mgf of a normal distribution, then you immediately know that this random variable must have a normal density, without having to explicitly derive the pdf from the mgf.
As an example, if $\left(N_t\right)$ is a Poisson process, then I can derive the fact that the time $T_1$ of the first arrival is exponentially distributed in the following way:
$$P\left(T_1 > t\right) = P\left(N_t = 0\right) = e^{-\lambda t}.$$
To get the final equality I am only using the fact that $N_t\sim Poisson\left(\lambda t\right)$. But through the first equality I have found that $T_1$ has the same cdf as an exponential distribution with rate $\lambda$. Since the cdf completely characterizes the distribution, I now automatically know the density of $T_1$.
Best Answer
HINT
Assuming $y \ge 0$, we have $$ F_Y(y) = \mathbb{P}[ Y \le y] = \mathbb{P}\left[ e^X \le y\right] = \mathbb{P}[ X \le \ln(y) ] $$ Can you finish this to find $F_Y(y)$ and differentiate to get $f_Y(y)$?