$\displaystyle \frac{x^2-y^2}{4xy-y^2} \cdot \frac{y-4x}{x^2+xy}$
$\displaystyle = \frac{(x-y)(x+y)}{y(4x-y)} \cdot \frac{y-4x}{x(x+y)}$
I'll break this into pieces so you'll see what is happening:
- $\displaystyle x^2 - y^2 = (x-y)(x+y)$. This is what the difference of squares is (ring a bell?).
- $\displaystyle 4xy - y^2 = y(4x - y)$. The most you can take out from the whole expression (i.e. BOTH terms) is a single $\displaystyle y$. You can't factor out an $\displaystyle x$ because the second term does not have one and you can't factor out another $\displaystyle y$ because the first term only has one $\displaystyle y$.
- $\displaystyle y-4x$. Not much we can do here but notice how this looks like $\displaystyle 4x - y$ in the second bullet above? They look similar but are not equal so that you can not directly cancel them. However, (remember this trick), if you factor out a -1 from this expression, we get: $\displaystyle -(4x - y)$ which we can now cancel with the other (4x - y).
- $\displaystyle x^2 + xy = x(x+y)$. The most you can factor out from both terms is an $\displaystyle x$. The reasoning is the same as the second bullet.
Now, see what you can cancel and hopefully you'll be able to do your other problems similarly as well.