$\begin{array}{l}\text{ Let }y=(mx+c)\text{ is tangent to }{x}^2=6y\\ solve the above equations \Rightarrow x^2=6(mx+c)\\ \text{ So, }{x}^2-6mx-6c=0\\ \text{ Put }D=b^2-4ac=0\\ \Rightarrow c=\frac{-3}{2}{m}^2\\ \therefore \text{ we get }y=mx-\frac{3}{2}{m}^2\dots \dots .\left(1\right)\end{array}$

$\begin{array}{l}\text{ And given hyperbola equation is }2x^2-4y^2=9\\ \Rightarrow \frac{x^2}{\frac{9}{2}}-\frac{y^2}{\frac{9}{4}}=1\dots .\left(2\right)\end{array}$

$\text{ Since, equation (1) is a tangent of equation (2) then }{c}^2=a^2{m}^2-b^2$

$\begin{array}{l}\Rightarrow \frac{9}{4}{m}^4=\frac{9}{2}{m}^2-\frac{9}{4}\\ \Rightarrow m^4=2m^2-1\\ \Rightarrow m^4-2m^2+1=0\\ \Rightarrow {\left(m^2-1\right)}^2=0\\ \Rightarrow m=\pm 1\\ \text{ for }m=1\text{ , equation of tangent is }x-y=\frac{3}{2}\end{array}$

Therefore, the correct answer is (D).