Find $f(x)$ where $f(x)(A-\frac{B}{x+B/A})+Cf(x+\frac{B}{A})=0$. $A, B, C > 0$, $x$ is complex and $Re(x)>0$. My guess is that $f(x)=0$ but I don't know how to prove it. 

For $f(x)\left(A-\dfrac{B}{x+C}\right)+Df(x+C)=0$ ,
$Df(x+C)=\dfrac{B-A(x+C)}{x+C}f(x)$
$f(Cx+C)=\dfrac{B-A(Cx+C)}{D(Cx+C)}f(Cx)$
$f(C(x+1))=\dfrac{B-AC(x+1)}{CD(x+1)}f(Cx)$
$f(C(x+1))=\dfrac{-AC\left(x-\dfrac{B}{AC}+1\right)}{CD(x+1)}f(Cx)$
$f(C(x+1))=-\dfrac{A}{D}\dfrac{x-\dfrac{B}{AC}+1}{x+1}f(Cx)$
With reference to http://eqworld.ipmnet.ru/en/solutions/fe/fe1105.pdf,
The general solution is $f(Cx)=\Theta_1(x)\dfrac{A^x\Gamma\left(x-\dfrac{B}{AC}+1\right)}{D^x\Gamma(x+1)}$, where $\Theta_1(x)$ is an arbitrary unit antiperiodic function
$f(x)=\Theta(x)\dfrac{A^\frac{x}{C}\Gamma\left(\dfrac{x}{C}-\dfrac{B}{AC}+1\right)}{D^\frac{x}{C}\Gamma\left(\dfrac{x}{C}+1\right)}$, where $\Theta(x)$ is an arbitrary antiperiodic function with period $C$
Similarly, for $f(x)\left(A-\dfrac{B}{x+\dfrac{B}{A}}\right)+Cf\left(x+\dfrac{B}{A}\right)=0$ ,
The general solution is $f(x)=\Theta(x)\dfrac{A^\frac{Ax}{B}\Gamma\left(\dfrac{Ax}{B}\right)}{C^\frac{Ax}{B}\Gamma\left(\dfrac{Ax}{B}+1\right)}=\Theta(x)\dfrac{A^{\frac{Ax}{B}-1}B}{C^\frac{Ax}{B}x}$, where $\Theta(x)$ is an arbitrary antiperiodic function with period $\dfrac{B}{A}$
