Multiplying the denominator $$\frac{2}{3} \gt -4y - \frac{25}{3}$$
My question is to solve this problem we need to multiply both sides by $3$. The result will be 
$$2 \gt -12y - 25.$$
Why doesn't the numerator of both $\frac{2}{3}$ and $\frac{25}{3}$ get multiplied?

As an expansion of what I said in the comments:
When we multiply the equation $\frac{2}{3} > -4y - \frac{25}{3}$  by $3$ we get the following equation
$\frac{2\times3}{3} > (-4y)\times3 - \frac{25\times3}{3}$. Simplifying the following we get
$\frac{6}{3} > -12y - \frac{75}{3}$.
Now $\frac{6}{3} =2$ and $\frac{75}{3} = 25$. Thus utilizing these known facts we can simplify the above equation giving us
$2 > -12y - 25$. So the numerators got multiplied by $3$, but then we divided by $3$, which amounts to returning to the original numerator. This is because multiplication and division are inverse operators. 
From your comment, if you changed the $3$ in the denominator to a $6$ you would not get the same result. Instead of ending up with $25$, you would get $\frac{25}{2}$.
