[tex]-\dfrac{7xy}{3x}+\dfrac{4y^2}{2y}=-\dfrac{7}{3}\cdot\dfrac{x}{x}\cdot y+\dfrac{4}{2}\cdot\dfrac{y^2}{y}\\\\\text{canceled x and one y}\\\\=-\dfrac{7}{3}y+\dfrac{4}{2}y=-3.5y+2y=-1.5y[/tex]
