We have that
$$
r_a(x) = {{\Gamma \left( {\left( {a + 1} \right)x} \right)} \over {\Gamma \left( x \right)}}
= {{\Gamma \left( {x + a\,x} \right)} \over {\Gamma \left( x \right)}} = x^{\,\overline {\,a\,x\,} }
$$
where $x^{\,\overline {\,y\,} } $ denotes the Rising Factorial.
Now, the Rising Factorial is defined (for $x$ and $y$ real and also complex) as
$$
h(x,y) = x^{\,\overline {\,y\,\,} } = {{\Gamma \left( {x + y} \right)} \over {\Gamma \left( x \right)}} = \prod\nolimits_{\;k\, = \,\,0}^{\,y} {\left( {x + k} \right)}
$$
where the last term denotes the Indefinite Product, computed for $k$ ranging between the indicated bounds.
So we can write $f_a(x)$ as
$$ \bbox[lightyellow] {
f_{\,a} (x) = 2^{\,a\,x} {{\Gamma \left( {x + ax} \right)} \over {\Gamma \left( x \right)}}
= \prod\nolimits_{\;k\, = \,\,0}^{\,a\,x} {2\left( {x + k} \right)} = 2^{\,a\,x} x^{\,a\,x} \prod\nolimits_{\;k\, = \,\,0}^{\,a\,x} {\left( {1 + k/x} \right)}
} \tag{1}$$
Concerning the derivative of $r_a(x)$, since
$$
\left\{ \matrix{
{\partial \over {\partial x}}h(x,y) = {{\Gamma \left( {x + y} \right)} \over {\Gamma \left( x \right)}}\left( {\psi \left( {x + y} \right)
- \psi \left( x \right)} \right) \hfill \cr
{\partial \over {\partial y}}h(x,y) = {{\Gamma \left( {x + y} \right)} \over {\Gamma \left( x \right)}}\psi \left( {x + y} \right) \hfill \cr} \right.
$$
then, as you already found
$$
\eqalign{
& {d \over {dx}}r_a(x) = {\partial \over {\partial x}}h(x,y) + {\partial \over {\partial y}}h(x,y){d \over {dx}}y = \cr
& = {{\Gamma \left( {x + ax} \right)} \over {\Gamma \left( x \right)}}\left( {\left( {a + 1} \right)\psi \left( {x + ax} \right)
- \psi \left( x \right)} \right) = \cr
& = {{\Gamma \left( {x + ax} \right)} \over {\Gamma \left( x \right)}}\left( {a\psi \left( {x + ax} \right) + \left( {\psi \left( {x + ax} \right)
- \psi \left( x \right)} \right)} \right) \cr}
$$
where, for $0<x$ (and $0<a$) $\Gamma(x+ax)/\Gamma(x)$ is clearly positive.
However, while $\psi(x+ax)-\psi(x)$ is also positive since $\psi(x)$ is increasing in that range, $a\psi(a+ax)$ introduces a negative term for lower $x$.
To determine the limit of $r_a'(x)$ as $x \to 0^+$, let's consider the series development of
$$ \bbox[lightyellow] {
\left\{ \matrix{
\ln \Gamma (cx) = \ln \left( {{1 \over {cx}}} \right) - \gamma cx + O\left( {x^{\,2} } \right) \hfill \cr
\psi (cx) = - {1 \over {cx}} - \gamma + {{\pi ^{\,2} } \over 6}cx + O\left( {x^{\,2} } \right) = \hfill \cr
= - {1 \over {cx}} - \gamma + \sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}} - {1 \over {k + 1 + cx}}} \right)} \hfill \cr} \right.
} \tag{2}$$
Therefore
$$
\eqalign{
& \mathop {\lim }\limits_{x\; \to \;0^{\, + } } {d \over {dx}}r_a(x) = \mathop {\lim }\limits_{x\; \to \;0^{\, + } } {{\Gamma \left( {x + ax} \right)} \over {\Gamma \left( x \right)}}\mathop {\lim }\limits_{x\; \to \;0^{\, + } } \left( {\left( {a + 1} \right)\psi \left( {x + ax} \right) - \psi \left( x \right)} \right) = \cr
& = {1 \over {a + 1}}\left( { - \gamma a} \right) \cr}
$$
which is negative for $0<a$.
Concerning $f_a(x)$ instead
$$
\eqalign{
& {d \over {dx}}f_{\,a} (x) = \;2^{\,a\,x} a\ln 2r_{\,a} (x) + 2^{\,a\,x} {d \over {dx}}r_{\,a} (x) = \cr
& = 2^{\,a\,x} r_{\,a} (x)\left( {a\ln 2 + {d \over {dx}}\ln \left( {r_{\,a} (x)} \right)} \right) = \cr
& = 2^{\,a\,x} r_{\,a} (x)\left( {a\ln 2 + {d \over {dx}}\ln \Gamma (x + ax) - {d \over {dx}}\ln \Gamma (x)} \right) = \cr
& = 2^{\,a\,x} r_{\,a} (x)\left( {a\ln 2 + \left( {a + 1} \right)\psi (x + ax) - \psi (x)} \right) \cr}
$$
(which is the equation you already found)
and
$$ \bbox[lightyellow] {
\eqalign{
& \mathop {\lim }\limits_{x\; \to \;0^{\, + } } f_{\,a} '(x) = 1\left( {\left( {a\ln 2} \right){1 \over {a + 1}} - {{\gamma a} \over {a + 1}}} \right) = \cr
& = {a \over {a + 1}}\left( {\ln 2 - \gamma } \right) = {a \over {a + 1}}0.1159 \cdots \cr}
} \tag{3}$$
Proceeding with the development of the derivative above
$$ \bbox[lightyellow] {
\eqalign{
& {d \over {dx}}f_{\,a} (x)\;\mathop /\limits_{} \;\left( {2^{\,a\,x} r_{\,a} (x)} \right) = \cr
& = a\ln 2 + \left( {a + 1} \right)\psi (x + ax) - \psi (x) = \cr
& = a\ln 2 + \left( {a + 1} \right)\left( { - {1 \over {\left( {a + 1} \right)x}} - \gamma + \sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}}
- {1 \over {k + 1 + \left( {a + 1} \right)x}}} \right)} } \right) - \left( { - {1 \over x} - \gamma + \sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}}
- {1 \over {k + 1 + x}}} \right)} } \right) = \cr
& = \left( {a\ln 2 - {1 \over x} - \left( {a + 1} \right)\gamma + {1 \over x} + \gamma } \right)
+ \left( {a + 1} \right)\sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}} - {1 \over {k + 1 + \left( {a + 1} \right)x}}} \right)}
- \sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}} - {1 \over {k + 1 + x}}} \right)} = \cr
& = a\left( {\ln 2 - \gamma } \right) + \sum\limits_{0\, \le \,k} {\left( {{a \over {k + 1}} + {1 \over {k + 1 + x}}
- {{\left( {a + 1} \right)} \over {k + 1 + \left( {a + 1} \right)x}}} \right)} = \cr
& = a\left( {\ln 2 - \gamma } \right) + a\sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}}
- {{k + 1} \over {\left( {k + 1 + x} \right)\left( {k + 1 + \left( {a + 1} \right)x} \right)}}} \right)} \cr}
} \tag{4}$$
and
$$ \bbox[lightyellow] {
\eqalign{
& 0 \le {1 \over {k + 1}} - {{k + 1} \over {\left( {k + 1 + x} \right)\left( {k + 1 + \left( {a + 1} \right)x} \right)}} = \cr
& = {1 \over {k + 1}} - {1 \over {\left( {1 + x/\left( {k + 1} \right)} \right)\left( {k + 1 + \left( {a + 1} \right)x} \right)}}\quad \left| {\;0 \le x,k} \right. \cr}
} \tag{5}$$
therefore your thesis is demonstrated.
Best Answer
Stirling's approximation says that $\log \Gamma(z) \sim z \log z + O(z)$. So $$\log \Gamma(\alpha x + 1)\sim (\alpha x + 1)\log(\alpha x + 1)+O(x)\sim \alpha x \log x + O(x),$$ and so certainly $\log f(x)\sim -\frac{1}{30}x\log x + O(x)$: your function must, eventually, start decreasing asymptotically to zero. But eventually is important. Let's try to estimate where this happens.
Consider the ratio $f(30y+30) / f(30y)$, where $y$ is an integer. You have $$ \frac{f(30y+30)}{f(30y)}=\frac{\Gamma(30y+31)\Gamma(15y+1)\Gamma(10y+1)\Gamma(6y+1)}{\Gamma(30y+1)\Gamma(15y+16)\Gamma(10y+11)\Gamma(6y+7)}=\frac{(30y+30)!(15y)!(10y)!(6y)!}{(30y)!(15y+15)!(10y+10)!(6y+6)!}=\frac{(30y+30)(30y+29)\cdots(30y+1)}{(15y+15)(15y+14)\cdots(15y+1)\cdot(10y+10)\cdots(10y+1)\cdot(6y+6)\cdots(6y+1)}\approx\frac{30^{30}}{15^{15}10^{10}6^{6}}\frac{1}{y}=\frac{1.008\times10^{12}}{y}, $$ where the $\approx$ is arrived at by pulling out the prefactors, counting powers of $y$, and ignoring the additive terms (which is justified when $y$ is large). This is going to be less than $1$ eventually... when $y$ exceeds a trillion. Your function will start decreasing when $x$ is greater than about $3\times 10^{13}.$ By that point, $f(x)$ will have many trillions of decimal digits, which I dare say is a bit more than Excel can handle.