The function $f$ satisfies
$$ \frac{f(x)}{f(y)} \le 2^{(x-y)^2}$$
for all $x,y$ in the domain of $f$. Then $f$ can be which of the following:
(A) $\sqrt{x}+x^3$
(B) $\int_0^{\sin^2 x} \sin^{-1}\sqrt{t}\,dt + \int_0^{\cos^2 x} \cos^{-1}\sqrt{t}\,dt$
(C) $\int_0^x 2t^3\,dt$
(D) $\frac{5}{2} x^2 + x^3 - 32$.
Suppose that our domain contain two different points $x$ and $y$.
Then we have $$ \frac{f(x)}{f(y)} = 2^{(x-y)^2} = 2^{(y-x)^2} = \frac{f(y)}{f(x)}.$$ Thus we see that $f(x)^2 = f(y)^2$ and thus $f(x) = \pm f(y)$, but then $$\pm 1 = \frac{f(x)}{f(y)} = 2^{(x-y)^2} \implies f(x) = f(y) \wedge (x-y)^2 = 0 \implies x=y.$$ Thus we see that there exists at most one point in our domain.
For this point $x$ we know only that $f(x) \neq 0$.
If instead of an equality, we have an inequality we can try to eliminate some of the answers.
For (A), we can consider $x=1$, $y=2$, then $\frac{f(y)}{f(x)} \ge \frac{8}{2} = 2^2 > 2 = 2^{(x-y)^2}$.
For (C), we can integrate and see that $f(x) = \frac{x^4}{2}$, so we use again $x=1$ and $y=2$ to obtain $\frac{f(y)}{f(x)} = \frac{16}{1} = 2^4 > 2 = 2^{(x-y)^2}$.
For (D), we can use $x=4$ and $y=5$ to obtain $\frac{f(y)}{f(x)} >2 = 2^{(x-y)^2}$.
So all that remains is (B), we take the domain of $x$ to be $[0, \frac{pi}{2}]$ for convienence. We all know from calculus the integral of $\arcsin(t)$ and of $\sqrt{1-t^2}$, so by integration by parts we obtain $$\int_0^{\sin(x)^2} \arcsin (\sqrt{t} ) \textrm{d} t = 2 \int_0^{\sin(x)} z \arcsin(z) \textrm{d} z = \frac{1}{2} \cos(x) \sin(x) + \sin(x)^2x -\frac{1}{2} x. $$ Likewise, we obtain $$\int_0^{\cos(x)^2} \arccos(\sqrt{t} ) \textrm{d} t =- \frac{1}{2} \sin(x) \cos(x) + \cos(x)^2 x + \arcsin(\cos(x)).$$ Thus we have $$ f(x) = \frac{1}{2} x + \arcsin(\cos(x))$$ And then try $x=\frac{\pi}{4}$ and $y=\frac{\pi}{3}$ to obtain $$ \frac{f(x)}{f(y)} = \frac{ \frac{1}{8}+ \frac{1}{4}}{\frac{1}{6}+ \frac{1}{6}}= \frac{9}{8} > 2^{ \frac{\pi^2}{144}} = 2^{(x-y)^2}.$$
So it looks like neither (A), nor (B), nor (C), nor (D) fulfills the functional equation.