Exponent rules- basic algebra Can some one point out where I've gone wrong. The correct answer, aside from the 9, is on the left page. I tried it again, changing up what I used as the denominator and it looks even more incorrect. Was what I did on the right page wrong?

 A: Your page on the right is incorrect for the reason I pointed out in my comment. 
Your page on the left is incorrect because in your second line you have taken the three from the denominator of your denominator into the numerator of your denominator for no reason.
A: We have: $\dfrac{\dfrac{x^{2}y^{-3}}{3z^{2}}-\dfrac{z^{-3}y^{-3}}{3x^{2}}}{\dfrac{x^{-4}y^{2}}{3z^{-2}}}$
$=\bigg(\dfrac{x^{2}y^{-3}}{3z^{2}}-\dfrac{z^{-3}y^{-3}}{3x^{2}}\bigg)\cdot\dfrac{3z^{-2}}{x^{-4}y^{2}}$
$=\bigg(\dfrac{(3x^{2})(x^{2}y^{-3})-(3z^{2})(z^{-3}y^{-3})}{(3z^{2})(3x^{2})}\bigg)\cdot\dfrac{3z^{-2}}{x^{-4}y^{2}}$
$=\bigg(\dfrac{3x^{4}y^{-3}-3y^{-3}z^{-1}}{9x^{2}z^{2}}\bigg)\cdot\dfrac{\dfrac{3}{z^{2}}}{\dfrac{y^{2}}{x^{4}}}$
$=\left(\dfrac{\dfrac{3x^{4}}{y^{3}}-\dfrac{3}{y^{3}z}}{9x^{2}z^{2}}\right)\cdot\dfrac{3}{z^{2}}\cdot\dfrac{x^{4}}{y^{2}}$
$=\left(\dfrac{\dfrac{(y^{3}z)(3x^{4})-(y^{3})(3)}{(y^{3})(y^{3}z)}}{9x^{2}z^{2}}\right)\cdot\dfrac{3x^{4}}{y^{2}z^{2}}$
$=\left(\dfrac{\dfrac{3x^{4}y^{3}z-3y^{3}}{y^{6}z}}{9x^{2}z^{2}}\right)\cdot\dfrac{3x^{4}}{y^{2}z^{2}}$
$=\dfrac{3x^{4}y^{3}z-3y^{3}}{9x^{2}y^{6}z^{3}}\cdot\dfrac{3x^{4}}{y^{2}z^{2}}$
$=\dfrac{9x^{8}y^{3}z-9x^{4}y^{3}}{9x^{2}y^{8}z^{5}}$
$=\dfrac{9x^{4}y^{3}\big(x^{4}z-1\big)}{9x^{2}y^{8}z^{5}}$
$=\dfrac{x^{2}\big(x^{4}z-1\big)}{y^{5}z^{5}}$
