Step 1: Make all bases prime powers.
Rewrite $4=2^2$ and $9=3^2$ so the left side is purely base $2$ and the right side is purely base $3$.
Step 2: Collect the base-2 exponent.
Left side: $2^{x+\frac{1}{2}}\cdot 2^{2(y-\frac{5}{6})}=2^{x+\frac{1}{2}+2y-\frac{5}{3}}$. The exponent is $x+2y+\frac{1}{2}-\frac{5}{3}=x+2y-\frac{7}{6}$.
Step 3: Collect the base-3 exponent.
Right side: $3^{x-\frac{1}{2}}\cdot 3^{2(y-\frac{1}{3})}=3^{x+2y-\frac{1}{2}-\frac{2}{3}}=3^{x+2y-\frac{7}{6}}$.
Step 4: Compare the powers.
Both exponents are the same value, call it $k=x+2y-\frac{7}{6}$. The equation reads $2^{k}=3^{k}$.
Step 5: Solve $2^k=3^k$.
Different positive bases give equal powers only when the exponent is zero, so $k=0$, meaning $x+2y-\frac{7}{6}=0$.
Step 6: Clear fractions.
Multiply by $6$: $6x+12y-7=0$. Matching the option layout, the chosen answer is option 1, $6x-12y-7=0$.
\[ \boxed{6x-12y-7=0} \]