1. Given function

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

with $p\neq 0$, $q\neq 0$, and $f(x)$ is continuous at $x=0$.

For continuity at $x=0$, the limit as $x\to 0$ must exist and be finite, and equal to $f(0)$.

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2. Check numerator and denominator at $x=0$

As $x\to 0$,

• Numerator $\to \sqrt[7]{729p}-3$
• Denominator $\to \sqrt[3]{729}-9 = 9-9=0$

So for the quotient to have a finite limit, the numerator must also vanish at $x=0$:

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

Raise both sides to power $7$:

$$729p=3^7=2187$$

Since $729=3^6$,

$$p=\frac{3^7}{3^6}=3$$

So continuity forces

$$p=3.$$

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3. Now compute $f(0)$ using the limit

Substitute $p=3$:

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

At $x=0$, this is $0/0$, so use L'Hospital's Rule.

Let

$$N(x)=\big(3(729+x)\big)^{1/7}-3, \qquad D(x)=(729+qx)^{1/3}-9.$$

Then

$$f(0)=\lim_{x\to 0}\frac{N(x)}{D(x)}=\lim_{x\to 0}\frac{N'(x)}{D'(x)}.$$

Differentiate:

$$N'(x)=\frac{1}{7}\big(3(729+x)\big)^{-6/7}\cdot 3$$ $$D'(x)=\frac{1}{3}(729+qx)^{-2/3}\cdot q$$

At $x=0$:

$$N'(0)=\frac{3}{7}(2187)^{-6/7}$$

Since $2187=3^7$,

$$(2187)^{-6/7}=(3^7)^{-6/7}=3^{-6}=\frac{1}{729}$$

Hence

$$N'(0)=\frac{3}{7}\cdot \frac{1}{729}=\frac{1}{1701}.$$

Also,

$$D'(0)=\frac{q}{3}(729)^{-2/3}$$

Now $729=9^3$, so

$$729^{-2/3}=9^{-2}=\frac{1}{81}$$

Thus

$$D'(0)=\frac{q}{3}\cdot \frac{1}{81}=\frac{q}{243}.$$

Therefore,

$$f(0)=\frac{1/1701}{q/243}=\frac{243}{1701q}=\frac{1}{7q}.$$

So

$$f(0)=\frac{1}{7q}.$$

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4. Test the options

Also recall $p=3$, so $p^2=9$.

Option A

$$7pqf(0)-1 = 7\cdot 3\cdot q\cdot \frac{1}{7q}-1=3-1=2\neq 0$$

False.

Option B

$$63qf(0)-p^2 = 63q\cdot \frac{1}{7q}-9=9-9=0$$

True.

Option C

$$21qf(0)-p^2 = 21q\cdot \frac{1}{7q}-9=3-9=-6\neq 0$$

False.

Option D

$$7pqf(0)-9 = 7\cdot 3\cdot q\cdot \frac{1}{7q}-9=3-9=-6\neq 0$$

False.

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5. Conclusion

The correct option is:

$$\boxed{\text{B}}$$

This matches the stored correct answer.