If $A$ is convex and $f$ has first partial derivative zero, then $f(x',y) = f(x,y)$ for every $(x',y),(x,y) \in A$? 
Let $A \subset \mathbb{R}^{2}$ be open and $f:A \to \mathbb{R}$ of class $C^{1}$. Suppose that $D_{1}f(a) = 0$ for every $a \in A$.
(a) If $A$ is convex, then $f(x',y) = f(x,y)$ for every $(x',y),(x,y) \in A$?
(b) And if we change "convex" by "connected"?

First, in the question it's really written $D_{1}f(a)$, but according to the hypothesis, I think it should be $D_{x}f(a)$, right?
For (a). I guess it's true, since we can connect $(x',y)$ to $(x,y)$ by a line contained in $A$ and by hypothesis, the variation on the coordinate $x$ is null. But, I cannot write this strictly
For (b). If $A$ is connected, we can connect $(x',y)$ to $(x,y)$ by a poligonal? And if yes, we can applied the item (a) in each line segmente of the poligonal?
 A: Notation. $D_1f(a)$ is the intended notation here. It means the derivative of $f$ with respect to the first variable. This notation is preferable to $D_x$ when the first variable isn't consistently called $x$. For example, here we see $x$ and $x'$ there. And when the arguments are written as a vector $a$ instead of individual components, $D_x(a)$ is really unclear: by looking at this formula, one doesn't know what $x$ has to do with $a$. 
You may also want to note the difference between, say, $D_1f(2x, 3y)$ and $D_xf(2x, 3y)$: specifically, $D_xf(2x, 3y) = 2D_1f(2x, 3y)$. 
(a) Fix $y$. Consider the function $g(x) = f(x, y)$. By the assumption $g'$ is identically zero on the interval $\{x: (x, y)\in A\}$. Apply the mean value theorem to $g$ and conclude it is constant. 
(b) For general connected domains the statement no longer holds. As a counterexample, let $A=\mathbb{R}^2\setminus \{(0, y):y\ge0 \}$ - that is, $A$ is the plane with a vertical half-line removed. Define
$$
f(x, y) = \begin{cases} y^2,\quad &\text{if } x>0 \text{ and } y>0 \\ 0 \quad &\text{otherwise} \end{cases}
$$
This is a $C^1$ function; its gradient is 
$$
\nabla f(x, y) = \begin{cases} (0, 2y),\quad &\text{if } x>0 \text{ and } y>0 \\ (0, 0) \quad &\text{otherwise} \end{cases}
$$
which is continuous in $A$ (note the importance of removing that half-line from $A$). Clearly, $D_1f\equiv 0$ but $f(-1, 1) = 0 \ne 1 = f(1, 1)$. 
