Differential Calculus !

Given , $f(x).f^{'}(-x)=f(-x).f^{'}(x)$

$\frac{f^{'}(x)}{f(x)}$ $=$ $\frac{f^{'}(-x)}{f(-x)}$

Integrating Both The Sides We Get ,

$lnf(x)=-lnf(-x)+c$

$ln(f(x).f(-x))=c$

$f(x).f(-x)=c$

Given , $f(0)=3$

$f^{2}(0)=9=c$

$f(x).f(-x)=9$

$\begin{aligned} \frac d{dx} f(x)f(-x) & = f'(x)f(-x) - f(x)f'(-x) = 0 & \small \color{#3D99F6} \text{Since }f(x)f'(-x) = f'(x)f(-x) \\ \implies f(x)f(-x) & = \color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ f(0)f(0) & = \boxed 9 & \small \color{#3D99F6} \text{Putting }x=0 \end{aligned}$

Assume that the answer is constant for all $x$ and all possible $f$ .

Since $f'$ appears on both sides of the equation, a trivial solution can be obtained by setting $f'(x) = 0$ for all $x$ (i.e. $f$ is a constant function), making the equation equivalent to $0=0$ .

Then, since $f(0) = 3$ , $f(x)\times f(-x) = 3\times 3 = 9$