1. Given conditions

We have $p+q=3$ and $p^4+q^4=369.$

We need to find $\left(\frac{1}{p}+\frac{1}{q}\right)^{-2}.$

2. Express the required quantity in terms of $pq$

First, $\frac{1}{p}+\frac{1}{q}=\frac{p+q}{pq}=\frac{3}{pq}.$

Therefore, $\left(\frac{1}{p}+\frac{1}{q}\right)^{-2}=\left(\frac{3}{pq}\right)^{-2}=\left(\frac{pq}{3}\right)^2=\frac{(pq)^2}{9}.$

So we only need to find $pq$.

3. Use the identity for $p^4+q^4$

We know $p^2+q^2=(p+q)^2-2pq=9-2pq.$

Now, $p^4+q^4=(p^2+q^2)^2-2p^2q^2.$

Substitute $p^2+q^2=9-2pq$ and $p^2q^2=(pq)^2$: $369=(9-2pq)^2-2(pq)^2.$

Expand: $369=81-36pq+4(pq)^2-2(pq)^2,$ $369=81-36pq+2(pq)^2.$

Rearrange: $2(pq)^2-36pq-288=0.$

Divide by $2$: $(pq)^2-18pq-144=0.$

Let $x=pq$. Then $x^2-18x-144=0.$

Solve: $x=\frac{18\pm\sqrt{18^2+4\cdot144}}{2}=\frac{18\pm\sqrt{324+576}}{2}=\frac{18\pm30}{2}.$

Hence, $x=24 \quad \text{or} \quad x=-6.$

So, $pq=24 \quad \text{or} \quad pq=-6.$

4. Use reality of $p,q$

Since $p$ and $q$ are real and satisfy $t^2-(p+q)t+pq=0,$ that is, $t^2-3t+pq=0,$ its discriminant must be non-negative: $9-4pq\ge 0.$

• If $pq=24$, then $9-96=-87<0,$ impossible for real $p,q$.
• Hence only $pq=-6$ is valid.

5. Compute the required value

We found $\left(\frac{1}{p}+\frac{1}{q}\right)^{-2}=\frac{(pq)^2}{9}.$

Substitute $pq=-6$: $\frac{(-6)^2}{9}=\frac{36}{9}=4.$

6. Final answer

$\boxed{4}$

7. Comparison with stored correct answer

Stored correct answer = $4$.

Our derived answer also equals $4$, so they agree.