Let $I_1,...,I_n \subset \mathbb{R}$ be open intervals and $U := I_1 \times \cdot \cdot \cdot \times I_n \subset \mathbb{R^n}.$
Further, let $\text { }f:U \to \mathbb{R}$ be a function that is partially derivable on all points with
$$\frac{\partial f}{\partial x_j} (x) = 0 \ \ \ \forall {j \in \text{{1,...,n}}} \ \ \ \forall x \in U $$
How can I show, that $f$ is constant?
So far I got this:
We take two generic points $x = (x_1, ...., x_n)$ and $y = (y_1,....,y_n) \in U.$
We want to show that $f(x) = f(y).$
Therefore we define the points $a_0 := y$, $a_n := x$, as well as $a_k := (x_1,....,x_k, y_{k+1},....,y_n) \text { for } 1 \leq k \leq n - 1.$
Now (for $k\in \text{{1,...,n}})$ we look at the functions $f_k: [0,1] \to \mathbb{R}$ with $f_k(t) := f(a_{k-1} + t(x_k - y_k)e_k)$, where $e_k$ is the k-th unit vector in $\mathbb{R^n}$.
How can I show now, that $f'_k \equiv 0$ for all $k$, from which we can say that all functions $f_k$ are constant? So, $f(a_{k-1}) = f_k(0) = f_k(1) = f(a_k) \text{ for all } k \in$ {$1,...,n$}. I have to deduce now that $f$ is constant.
And how can one conclude from above, that $\arctan (\frac{x}{y}) + \arctan(\frac{y}{x}) = \frac{\pi}{2}$ for all $x,y > 0$?
I know that $\tan \frac{\pi}{4} = 1$, but I don't know how to conclude $\frac{\pi}{2}$ from what is given above.
For your second question, define $f(x,y) = \arctan \frac xy + \arctan \frac yx$ on the quarter-space $(0,\infty) \times (0,\infty)$. Then $$\frac{\partial f}{\partial x} = \frac{1}{1 + (\frac xy)^2} \cdot \frac 1y + \frac{1}{1 + (\frac yx)^2} \cdot \frac{-y}{x^2} = 0$$ after a bit of simplification. Since $f$ is symmetric in $x$ and $y$ you get also that $\dfrac{\partial f}{\partial y} = 0$, so that $f$ is constant.
It follows that $f(x,y) = f(1,1) = \dfrac \pi 2$ for all $x,y > 0$.