$\text{Given }, f\left(x\right)=\frac{3^x+3^{-x}}{2}$

$\therefore f\left(x+y\right)=\frac{3^{x+y}+3^{-(x+y)}}{2} \text{and} f\left(x-y\right)=\frac{3^{x-y}+3^{-(x-y)}}{2}$

Now equating the equations

$f(x+y)+f(x-y)=af\left(x\right)f\left(y\right)$

$\Rightarrow \frac{3^{x+y}+3^{-(x+y)}}{2}+\frac{3^{x-y}+3^{-(x-y)}}{2}=a\frac{3^x+3^{-x}}{2}\times \frac{3^y+3^{-y}}{2}$

$\Rightarrow \frac{\bcancel{3^{x+y}+3^{-(x+y)}+3^{x-y}+3^{-(x-y)}}}{2}=a\frac{\bcancel{3^{x+y}+3^{-(x+y)}+3^{x-y}+3^{-(x-y)}}}{4}$

$\Rightarrow a=2$