Find the value.

We use here the identity $a^{2}+b^{2}=(a+b)^{2}-2ab=(a-b)^{2}+2ab$ . Given here,

$\frac{a}{b}+\frac{9b}{a}=7$

$\implies \frac{a^{2}+9b^{2}}{ab}=7$ ....(i)

$\implies \frac{(a+3b)^{2}-6ab}{ab}=7$

$\implies \frac{(a+3b)^{2}}{ab}-6=7 \implies \frac{(a+3b)^{2}}{ab}=13 \implies (a+3b)^{2}=13ab$ ...(ii)

Also, we can rewrite (i) as,

$\implies \frac{a^{2}+9b^{2}}{ab}=7$

$\implies \frac{(a-3b)^{2}+6ab}{ab}=7$

$\implies \frac{(a-3b)^{2}}{ab}+6=7 \implies \frac{(a-3b)^{2}}{ab}=1 \implies (a-3b)^{2}=ab$ ...(iii)

From (ii) and (iii), we have,

$(\frac{a+3b}{a-3b})^{2}=\frac{(a+3b)^{2}}{(a-3b)^{2}}=\frac{13ab}{ab} = \boxed{13}$

$\frac{a}{b}+\frac{9b}{a}=\frac{a^2+(3b)^2}{ab}=7$ $\rightarrow$ $a^2+(3b)^2=7ab$ .Adding $6ab$ to both sides to complete the square and we get : $(a+3b)^2=13ab$ $\rightarrow$ $a+3b=\sqrt{13ab}$ .Similarly we have : $a^2+(-3b)^2=7ab$ .Substracting $6ab$ from both sides we get : $(a-3b)^2=ab$ $\rightarrow$ $a-3b=\sqrt{ab}$ , therefore $\frac{a+3b}{a-3b}=\frac{\sqrt{13ab}}{\sqrt{ab}}=\sqrt{13}$ and by squaring this value we get $\boxed{13}$ .