If for $p\ne q\ne 0$, then function $f\left(x\right)=\frac{\sqrt[7]{p\left(729+x\right)}-3}{\sqrt[3]{729+qx}-9}$ is continuous at $x=0$, then

For a function to be continuous $f\left(x\right)$ to be $x=0$, we know 

$f\left(0\right)=\underset{x\to 0}{\lim }f\left(x\right)$

For $f\left(x\right)=\frac{\sqrt[7]{p\left(729+x\right)}-3}{\sqrt[3]{729+qx}-9}$ 

limit should be $\frac{0}{0}$ from

So, $\underset{x\to 0}{\lim }\left(\sqrt[7]{p\left(729+x\right)}-3\right)$ 

$\sqrt[7]{p.729}-3=0$

$\Rightarrow p·729=3^7 \Rightarrow p=3$

Now, $f\left(0\right)=\underset{x\to 0}{\lim }\frac{\sqrt[7]{3\left(3^6+x\right)}-3}{\sqrt[3]{3^6+qx}-9}$

$=\underset{x\to 0}{\lim }\frac{3\left[{\left(1+\frac{x}{3^6}\right)}^{{\displaystyle \frac{1}{7}}}-1\right]}{9\left[{\left(1+\frac{qx}{3^6}\right)}^{{\displaystyle \frac{1}{3}}}-1\right]}=\frac{3}{9}\times \frac{\frac{1}{7·3^6}}{\frac{\mathrm{q}}{3·3^6}}$

$\Rightarrow f\left(0\right)=\frac{1}{3}\times \frac{3}{7q}=\frac{1}{7q}$

$\Rightarrow 7qf\left(0\right)-1=0$

$\Rightarrow 7·p^2·qf\left(0\right)-p^2=0$  (multiplying the equation by $p^2$)

$\Rightarrow 63qf\left(0\right)-p^2=0$