Answer
$${F_p}\left( {p,q} \right) = \frac{{2p + q}}{{2\sqrt {{p^2} + pq + {q^2}} }},\,\,\,\,\,\,\,\,\,{F_q}\left( {p,q} \right) = \frac{{p + 2q}}{{2\sqrt {{p^2} + pq + {q^2}} }}$$
Work Step by Step
$$\eqalign{
& F\left( {p,q} \right) = \sqrt {{p^2} + pq + {q^2}} \cr
& {\text{Find the partial derivatives }}{F_p}\left( {p,q} \right){\text{ and }}{F_q}\left( {p,q} \right){\text{ then}} \cr
& {F_p}\left( {p,q} \right) = \frac{\partial }{{\partial p}}\left[ {\sqrt {{p^2} + pq + {q^2}} } \right] \cr
& {\text{treat }}q{\text{ as a constant}} \cr
& {\text{use }}\left[ {\sqrt u } \right]' = \frac{{u'}}{{2\sqrt u }} \cr
& {F_p}\left( {p,q} \right) = \frac{1}{{2\sqrt {{p^2} + pq + {q^2}} }}\frac{\partial }{{\partial p}}\left[ {{p^2} + pq + {q^2}} \right] \cr
& {F_p}\left( {p,q} \right) = \frac{1}{{2\sqrt {{p^2} + pq + {q^2}} }}\left( {2p + q} \right) \cr
& {F_p}\left( {p,q} \right) = \frac{{2p + q}}{{2\sqrt {{p^2} + pq + {q^2}} }} \cr
& \cr
& and \cr
& \cr
& {F_q}\left( {p,q} \right) = \frac{\partial }{{\partial q}}\left[ {\sqrt {{p^2} + pq + {q^2}} } \right] \cr
& {\text{treat }}p{\text{ as a constant}} \cr
& {\text{use }}\left[ {\sqrt u } \right]' = \frac{{u'}}{{2\sqrt u }} \cr
& {F_q}\left( {p,q} \right) = \frac{1}{{2\sqrt {{p^2} + pq + {q^2}} }}\frac{\partial }{{\partial q}}\left[ {{p^2} + pq + {q^2}} \right] \cr
& {F_q}\left( {p,q} \right) = \frac{1}{{2\sqrt {{p^2} + pq + {q^2}} }}\left( {p + 2q} \right) \cr
& {F_q}\left( {p,q} \right) = \frac{{p + 2q}}{{2\sqrt {{p^2} + pq + {q^2}} }} \cr} $$
