Under what conditions can a function $ y: \mathbb{R} \to \mathbb{R} $ be expressed as $ \dfrac{z'}{z} $? Can an arbitrary function $ y: \mathbb{R} \to \mathbb{R} $ always be expressed as $ \dfrac{z'}{z} $ for some differentiable function $ z: \mathbb{R} \to \mathbb{R} $, or are additional conditions on $ y $ needed for this to be true? This question was inspired by reading a writeup about differential equations, where the change of variables $ y = \dfrac{z'}{z} $ is made.
 A: Let $ y: \mathbb{R} \to \mathbb{R} $ be any continuous function. Define $ z: \mathbb{R} \to \mathbb{R} $ by
$$
\forall x \in \mathbb{R}: \quad
z(x) \stackrel{\text{def}}{=} e^{F(x)},
$$
where $ F $ is an antiderivative of $ y $ (which exists). Notice that $ z $ is a positive function. By the Exponential Chain Rule,
$$
\forall x \in \mathbb{R}: \quad
z'(x) = e^{F(x)} \cdot y(x).
$$
Therefore,
$$
\forall x \in \mathbb{R}: \quad
y(x) = \dfrac{e^{F(x)} \cdot y(x)}{e^{F(x)}} = \dfrac{z'(x)}{z(x)}.
$$

Additional Set-Theoretic Information
Let $ y: \mathbb{R} \to \mathbb{R} $, and suppose that there exists a differentiable $ z: \mathbb{R} \to \mathbb{R}^{+} $ such that $ y = \dfrac{z'}{z} $. As $ \dfrac{z'}{z} = (\ln \circ z)' $, it follows that $ y $ is a derivative (or equivalently, $ y $ has an anti-derivative). Then according to the rather fantastic answer in this post, the set of discontinuities of $ y $ is a meager $ F_{\sigma} $-subset of $ \mathbb{R} $.
Conversely, if the set of discontinuities of $ y $ is a meager $ F_{\sigma} $-subset of $ \mathbb{R} $, then $ y $ has an anti-derivative $ F $, which can be used to construct the required function $ z $ in the manner above.

Conclusion: Let $ y: \mathbb{R} \to \mathbb{R} $. Then there exists a differentiable $ z: \mathbb{R} \to \mathbb{R}^{+} $ such that $ y = \dfrac{z'}{z} $ if and only if the set of discontinuities of $ y $ is a meager $ F_{\sigma} $-subset of $ \mathbb{R} $.

