$p + q =3 \quad p ^{4}+ q ^{4}=369$

$\left(\frac{1}{p}+\frac{1}{q}\right)^{-2}$

$(p+q)^{2}=9$

$p ^{2}+ q ^{2}=9-2 pq$

$\frac{1}{\left(\frac{1}{p}+\frac{1}{q}\right)^{2}}=\frac{(q p)^{2}}{(q+p)^{2}}=\frac{(q p)^{2}}{9}$

$p ^{4}+ q ^{4}=\left( p ^{2}+ q ^{2}\right)^{2}-2 p ^{2} q ^{2}$

$369=(9-2 p q)^{2}-2(p q)^{2}$

$369=81+4 p ^{2} q ^{2}-36 pq -2 p ^{2} q ^{2}$

$288=2 p ^{2} q ^{2}-36 pq$

$144= p ^{2} q ^{2}-18 pq$

$( pq )^{2}-2 \times 9 \times pq +9^{2}=144+9^{2}$

$(p q-9)^{2}=225$

$p q-9=\pm 15$

$p q=\pm 15+9$

$p q=24,-6$

($24$ is rejected because $p ^{2}+ q ^{2}=9-2 pq$ is negative)

$\frac{(q p)^{2}}{9}=\frac{16 \times 16}{9}=4$