What $Re(f(z))=c$ can be if $f$ is a holomorphic function ? Suppose that $f:U\subset\mathbb{C}\to\mathbb{C}$, where $U$ is a region in the complex plane, is a holomorphic function. 
Of course, if $c\in\mathbb{R}$ is a regular value for $\text{Re}(f(z))$ then $\text{Re}(f(z))^{-1}(c)$ is an union of differentiable curves in the plane. 
 Question: 
If $c$ is not a regular value and $\text{Re}(f(z))^{-1}(c)$ have at least one cluster point is this set also a piece-wise differentiable curve ?
 A: You are essentially asking for a description of the zero-set of a harmonic function in two variables. That is a very delicate issue, but has been studied a lot.
Maybe looking at:
L. De Carli, S. M. Hudson, Geometric remarks on the level curves of harmonic functions,
Bull. Lond. Math. Soc. 42 (2010), no. 1, 83–95.
L. Flatto, A theorem on level curves of harmonic functions, J. London Math. Soc. (2) 1 (1969) 470–472.
Z. Y. Wen, L. M. Wu, and Y. Zhang, Set of zeros of harmonic functions of two variables, Harmonic analysis, Tianjin, 1988, Lecture Notes in Mathematics 1494 (Springer, Berlin, 1991) 196–203
will help. I do not know the answer to your specific question.
A: Yes. The set of points in $\mathbb C$ having real part equal to $c$ form a line, i.e. a smooth simple curve $l$. The counterimage $f^{-1}(l)$ of any smooth simple curve $l$ via a holomorphic function $f$ is always piecewise smooth. 
To prove the last sentence, take any point $z_0\in U$ with $f(z_0) \in l$. There are two local diffeomorphisms at $z_0$ and $f(z_0)$ that move $z_0$ to $0$ and transform $f$ locally into $g(z) = z^n$. Since $f$ is not constant, we have $n>0$. Local diffeomorphisms send smooth curves to smooth curves. The counterimage along $g$ of a smooth curve passing through $0$ is the union of $n$ smooth curves exiting from 0. Therefore $f^{-1}(l)$ is a piecewise smooth curve.
