Question regarding non-linear (everywhere) differentiable functions symmetric wrt a line $p\ldots y=kx,k>0$ that also intersect that line 
Let $I\subset\Bbb R$ be an open interval and let $f(x):I\to\Bbb R$ be a non-linear (everywhere) differentiable function symmetric wrt a line $p\ldots y=kx,k>0$ that also intersects that line. Is $f(x)$ necessarily (a branch of) a rotated and possibly translated hyperbola $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ ?


Motivation:
At first, I was wondering whether there are unbounded continuous even functions whose graphs can be rotated around the origin so that the output is an injective function. This thought lead me to injective functions symmetric wrt some line passing through the origin. We could rotate those functions and get the wanted even functions.
The first example that came to my mind, other than a linear function which would give a constant, was a rational function $f(x)=\frac1x$, which might be a rotated hyperbola $\frac{y^2}2-\frac{x^2}2=1$.

Asymptotes of $f(x)=\frac1x$, the $x$ and $y$ axis, are perpendicular, so I took into account the upper branches of all the hyperbolas $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ in order to deal with the case of $p\ne y=x$, as in the picture (I went on that track because I believed one of the asymptotes would be horizontal, so if the asymptotes make an obtuse angle, that would prevent us from getting something that isn't a function):

I know the hyperbola I'm referring to is essentially just a rotated hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, but I wanted to make my starting point a little bit more practical.
Now, I was wondering whether such rotated branches were the only non-linear continuous injective functions symmetric to a line passing through the origin, but that was quite unlikely and I found a counter-example: $f(x)=\begin{cases}x^2,&x\leqslant 0\\-\sqrt{x},&x>0\end{cases}$.

I noticed it isn't differentiable at the origin because $\lim\limits_{x\to 0^-}2x=0$, while $\lim\limits_{x\to 0^+}-\frac1{2\sqrt{x}}=-\infty$.

This made me think about how to get rid of that spike at $x=0$, while preserving the differentiability on the rest of the domain. I thought the functions I could get by linear transformations of branches of hyperbolas might be the only ones satisfying that condition. However, I suspect my reasoning isn't quite right. So, may I ask:
Are there any counter-examples of such functions that:

*

*are symmetric wrt to a line $y=kx$

*intersect the same line at some point

*are differentiable at every point in their domain

*aren't linear

*don't belong to the family of the above-mentioned functions


Thank you very much in advance!
 A: Take any continuous function (not necessary derivable) meeting your requirements. Call if $f$.
Take a symmetric test function, i.e., a function with a compact support that is infinitely differentiable (example below); call it $g$.
Using a new x-axis that is perpendicular to the line of reflection, calculate the convolution $f*g$.
The resulting function is infinitely differentiable and meets your requirements.
Test function example:$g = e^{1/(x^2-1)} for |x| \lt 1, \quad 0  otherwise.$
A: I think your initial idea of rotating the graph of an even function works quite generally.
We can start with an arbitrary even function $f: \Bbb R \to \Bbb R$ which is differentiable with $|f'(x)| \le c < 1$ for all $x$. (For example, $f(x) = \frac 12 \cos(x)$). Now consider the equation
$$
 H(x, y) = x+y - f(x-y) = 0 \, .
$$
For each $x \in \Bbb R$ there is exactly one $y \in \Bbb R$ satisfying $H(x, y) = 0$, because for fixed $x$ the mapping
$$
 y \mapsto H(x, y) - y = x - f(x-y)
$$
is a (strict) contraction, so that it has exactly one fixed point. So there is a function $g: \Bbb R \to \Bbb R$ with $H(x, g(x)) = 0$ for all $x$.
The graph of $g$ is symmetric with respect to the line $y=x$ because $H(x, g(x)) = H(g(x), x)$, and intersects that line at $(x, y) = (\frac 12 f(0), \frac 12 f(0))$.
Finally, the differentiability of $g$ follows from the implicit function theorem, because
$$
 \frac{\partial H}{\partial y} (x, y) = 1+f'(x-y) \ge 1-c > 0
$$
