Growth of the gradient of $f(x+y) \leq f(x) f(y)$ Let $f: \mathbb{R}^3 \rightarrow \mathbb{R}_{\geq 0}$ be a radial continuous function and $C^2$ on $\mathbb{R}^3 \setminus \{0\}$ which satisfies the following functional inequality
$$ f(x+y) \leq f(x) f(y) $$
Does there exist constants $c,d$ such that for all $x\in \mathbb{R}^3 \setminus\{ 0\}$ we have
$$ \vert \nabla f (x) \vert \leq c e^{d\vert x\vert} \quad (1)$$
In other words

If $f$ solves $f(x+y) \leq f(x)f(y)$, does its gradient also grow at most exponentially? How about the Hessian?

In case we take $g(\vert x \vert)= f(x)$ and replace the inquality by an equality $g(r+s) = g(r) g(s)$ we get that $g(r) = e^{kr}$ for some $k$. Thus, in that case $f(x) = e^{k\vert x \vert}$. This function clearly satisfies the estimate $(1)$, as for $x\neq 0$
$$ \vert \nabla e^{k \vert x \vert} \vert = \left\vert \frac{kx}{\vert x \vert} e^{k\vert x \vert} \right\vert = \vert k \vert e^{k \vert x \vert}.$$
However, intuitively the function should be "less steep" when we take the functional inequality instead of the equality.
Added: In case it helps it would be fine if you assume $f\in C^\infty(\mathbb{R}^3)$.
 A: The answer to my question can essentially be found in the answer of another question here Derivatives of functions satisfying Euler-ish inequality $f(x+y)\le f(x)f(y)$.
Many thanks to @PhoemueX and @MaximilianJanisch (in particular for pointing out that we can actually smooth it without any problems).
Let me just give the main idea here. If we are in one dimension, then every function $f: \mathbb{R} \rightarrow \mathbb{R}$ with $3 e^x \leq f(x) \leq 5 e^x$ solves the functional inequality as
$$ f(x+ y) \leq 5 e^{x+y} < (3 e^x)(3e^y) \leq f(x) f(y).  $$
(More generally: Any function $f$ that satisfies $c\cdot e^x\le f(x)\le c^2\cdot e^x$ for a constant $c\geq 1$ satisfies the functional inequality.)
Now we apply this idea to my problem. Hence, we pick a function $f$ as above, then $h(x):= f(\vert x \vert)$ will be radial and
$$ h(x+y) = f(\vert x +y \vert ) \leq 5 e^{\vert x + y \vert} < (3 e^{\vert x \vert})(3 e^{\vert y \vert}) \leq h(x) h(y). $$
We could almost pick the example from the other answer, namely
$$ f(x) = e^x (4 + \sin(e^{x^2})) $$
we would just need to smooth it slightely at the origin. For example
$$ g(x) = e^{\vert x \vert + (\vert x \vert^2 - \vert x \vert) \chi(\vert x \vert)} (4 + \sin(e^{\vert x \vert^2})) $$ 
with a cut-off function supported around the origin will work. In fact this function is smooth and alread the gradient will behave terribly. Of course we could make it worse by replacing $e^{\vert x \vert^2}$ in the sinus by something which grows even faster. This tells us that our gradient will not just be not bounded by some exponential, but we can make it arbitrarily bad. 
Added: And, as Maximilian pointed out in the comment below, it works in any positive dimension. 
