Does there exist a real function with following properties?
As a physicist by education, I cannot make complete sense of your question. The main problem encountered at first sight
is: dimensionality.
Suppose, for example, that the function $f(x)$ in your question is a position $f$ dependent on time $x$,
replacing $t$ (what's in a name) with dimension $T$, and $f$ has the dimension of length $L$.
Then the derivative of this position with respect to time is $df/dx$, a velocity, and it has dimension $LT^{−1}$.
The second derivative $d^2f/dx^2 = d(df/dx)/dx$, an acceleration, has dimension $LT^{−2}$.
Now the sum in (3.) tries to add an acceleration to a velocity squared: $f''(x) + [f'(x)]^2$.
This may be correct mathematically, but it is physically impossible.
What is physically possible is to compare the general acceleration $f''(x)$
with a special acceleration, namely the one belonging to a
Centripetal force:
$a_c = v^2/r$. Translated into your parlance: $a_c = [f'(x)]^2/f(x)$.
Thus the following modification of your statement (3.) might make sense physically.
$$
f''(x)f(x) + [f'(x)]^2 < 0 \quad \mbox{with} \quad [f^2(x)]'' = [2f(x)f'(x)]' = 2f''(x)f(x) + 2[f'(x)]^2
$$
Which doesn't mean that a solution does indeed exist, though. What follows is my best shot. The (1.) whole domain $\mathbb{R}$
remains problematic in the partial solution below.
$$
f(x) = \sqrt{x\left[\frac{\pi}{2}-\arctan(x)\right]+\ln(1+x^2)/2} \\
f'(x) = \frac{\pi/2-\arctan(x)}{2\sqrt{x\left[\pi/2-\arctan(x)\right]+\ln(1+x^2)/2}} > 0 \quad \mbox{: only for} \; x > 0 \\
[f^2(x)]'' = \left[\frac{\pi}{2}-\arctan(x)\right]' = - \frac{1}{1+x^2} < 0
$$
One can "delay" the domain error $x > 0$ by augmenting a large constant $C > 0$ in the function $f(x)$, but that doesn't help in the end:
$$
f(x) = \sqrt{x\left[\frac{\pi}{2}-\arctan(x)\right]+\ln(1+x^2)/2+C}
$$
Any better ideas?