Correct. Using a common denominator of $\displaystyle{21}{{m}}^{{2}}{{n}}^{{4}}\usepackage{cmbright}\Huge\$ , you found that $\displaystyle\frac{{{6}{m}}}{{{7}{{m}}^{{2}}{n}}}-\frac{{{5}{{n}}^{{3}}}}{{{3}{m}{{n}}^{{4}}}}=\frac{{{18}{m}{{n}}^{{3}}-{35}{m}{{n}}^{{3}}}}{{{21}{{m}}^{{2}}{{n}}^{{4}}}}=\frac{{-{17}{m}{{n}}^{{3}}}}{{{21}{{m}}^{{2}}{{n}}^{{4}}}}=\frac{{-{17}}}{{{21}{m}{n}}}\usepackage{cmbright}\Huge\$ .