$\displaystyle 3x+ \frac 1 {2y}=9 \\ \displaystyle 9x^2+\frac 1 {4y^2}+2 \cdot 3x \frac 1 {2y}=81 \\ \displaystyle 9x^2+ \frac 1 {4y^2}+3 \frac x y=81 \\ \displaystyle 9x^2+ \frac 1 {4y^2}+45=81 \\ \displaystyle 9x^2+ \frac 1 {4y^2}=36 \\ \displaystyle \frac 4 9 (9x^2+ \frac 1 {4y^2})=\frac 4 9 \cdot 36 \\ \displaystyle 4x^2+ \frac 1 {9y^2}=\fbox{16}$