Yes, you are on the right track, but in order to do that you will need to multiply up by the denominators, or their multiples, to get the lowest common denominator, I get $4xy$.
Multiply the left fraction by $\frac{2x}{2x}$ and the right fraction by $\frac{y}{y}$
$\frac{3x}{2y}(\frac{2x}{2x})-\frac{7y}{4x}(\frac{y}{y})=\frac{6x^2}{4xy}-\frac{7y^2}{4xy}=\frac{6x^2-7y^2}{4xy}$
As you have stated in the comments that the actual question is: $\frac{3x}{2y-\frac{7y}{4x}}$, if I have understood you correctly, I will answer that now:
$\frac{3x}{2y(\frac{4x}{4x})-\frac{7y}{4x}}=\frac{3x}{\frac{2y-5y}{4x}}=\frac{3x(4x)}{-5y}=\frac{-12x^2}{5y}$