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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?

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    On the right hand page on the second line down in your second simplification, you have gotten one of the terms in the numerator wrong. There should be an $x^{-2}$ in the numerator of one of the terms in your numerator, not $x^2$ – 2017-01-08
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    Ohhh, the term on the right. But the 9 in the denominator of the left page, is that rightly there? The book has the answer without it. – 2017-01-08
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    See my answer below. – 2017-01-08

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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.

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We have: $\dfrac{\dfrac{x^{2}y^{-3}}{3z^{2}}-\dfrac{z^{-3}y^{-3}}{3x^{2}}}{\dfrac{x^{-4}y^{2}}{3z^{-2}}}$

$=\bigg(\dfrac{x^{2}y^{-3}}{3z^{2}}-\dfrac{z^{-3}y^{-3}}{3x^{2}}\bigg)\cdot\dfrac{3z^{-2}}{x^{-4}y^{2}}$

$=\bigg(\dfrac{(3x^{2})(x^{2}y^{-3})-(3z^{2})(z^{-3}y^{-3})}{(3z^{2})(3x^{2})}\bigg)\cdot\dfrac{3z^{-2}}{x^{-4}y^{2}}$

$=\bigg(\dfrac{3x^{4}y^{-3}-3y^{-3}z^{-1}}{9x^{2}z^{2}}\bigg)\cdot\dfrac{\dfrac{3}{z^{2}}}{\dfrac{y^{2}}{x^{4}}}$

$=\left(\dfrac{\dfrac{3x^{4}}{y^{3}}-\dfrac{3}{y^{3}z}}{9x^{2}z^{2}}\right)\cdot\dfrac{3}{z^{2}}\cdot\dfrac{x^{4}}{y^{2}}$

$=\left(\dfrac{\dfrac{(y^{3}z)(3x^{4})-(y^{3})(3)}{(y^{3})(y^{3}z)}}{9x^{2}z^{2}}\right)\cdot\dfrac{3x^{4}}{y^{2}z^{2}}$

$=\left(\dfrac{\dfrac{3x^{4}y^{3}z-3y^{3}}{y^{6}z}}{9x^{2}z^{2}}\right)\cdot\dfrac{3x^{4}}{y^{2}z^{2}}$

$=\dfrac{3x^{4}y^{3}z-3y^{3}}{9x^{2}y^{6}z^{3}}\cdot\dfrac{3x^{4}}{y^{2}z^{2}}$

$=\dfrac{9x^{8}y^{3}z-9x^{4}y^{3}}{9x^{2}y^{8}z^{5}}$

$=\dfrac{9x^{4}y^{3}\big(x^{4}z-1\big)}{9x^{2}y^{8}z^{5}}$

$=\dfrac{x^{2}\big(x^{4}z-1\big)}{y^{5}z^{5}}$