Multiply.

Thanks Aziz for that great explanation.

I just thought i would show how the cancelling is done in fraction form.

(I am playing with LaTex,  some of this is a bit new for me  )

$$\dfrac{9x^5y^4}{4y}\times\dfrac{2y}{3x}\\\\\\
=\dfrac{\not{9} \textcolor[rgb]{0,0,1}{^3} x^{\not5 \textcolor[rgb]{0,0,1}{4}} y^4}{\not{4}\textcolor[rgb]{0,0,1}{^2}\not{y}}\times\dfrac{\not{2}\not{y}}{\not{3}\not{x}}\\\\\\\
=\dfrac{3x^4y^4}{2}$$

Compare the cross or diagonal terms first.

9x^5*y^4 & 3x --> we can cross out the 3x entirely and change the 9x^5*y^4 into 3x^4*y^4 by dividing 9 by 3 and crossing out one x in x^5.

4y & 2y --> we can change 4y into 2(2y) and cross out the 2y.

Notice by crossing out the 3x and 2y, we are left with 1 / 1.

Thus, we have: 3x^4*y^4 / 2  * 1 / 1 -->

3x^4*y^4 / 2.