The easy temptation with a problem like this is to start expanding everything out. For this problem, this will quickly become a giant mess. Instead, we have to recognize a few things that will make the problem signifcantly easier. First, we should see that there are several terms being multiplied by (x-3z) and several are being multiplied by (3z-x). Let's pull those terms out and see how it looks:

 

 ((2x^2(x-3z)-5x(x-3z)+2(x-3z))/(4x^2(3z-x)-11x(3z-x)+6(3z-x)))/((4x+1)/(4x-3))

becomes: 

(((x-3z)(2x^2 - 5x + 2))/((3z-x)(4x^2 - 11x + 6)))/((4x+1)/(4x-3))

 

Now that doesn't look a whole lot better yet, but we are making progress. Secondly, we can examine ther terms (2x^2 - 5x + 2) and (4x^2 - 11x + 6). They can probably be factored. 

(2x^2 - 5x + 2) is factored: (2x -1)(x -2)

 

 (4x^2 - 11x + 6) is factored: (4x - 3)(x -2)

 

Now let's rewrite with this new simplification: 

(((x - 3z)(2x - 1)(x - 2))/((3z-x)(4x - 3)(x -2)))/((4x+1)/(4x-3))

Let's rewrite it vertically so that we can see it properly: 

$${\frac{\left({\frac{\left(\left(\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{z}}\right){\mathtt{\,\times\,}}\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}\right)\right){\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{2}}\right)\right)}{\left(\left(\left({\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{z}}{\mathtt{\,-\,}}{\mathtt{x}}\right){\mathtt{\,\times\,}}\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right)\right){\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{2}}\right)\right)}}\right)}{\left({\frac{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}}\right)}}$$

We can see an (x-2) term in both parts of the numerator of the fraction, this means that we can cancel that term: 

$${\frac{\left({\frac{\left(\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{z}}\right){\mathtt{\,\times\,}}\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}\right)\right)}{\left(\left({\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{z}}{\mathtt{\,-\,}}{\mathtt{x}}\right){\mathtt{\,\times\,}}\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right)\right)}}\right)}{\left({\frac{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}}\right)}}$$

Now we know when dividing by fractions, we can multiply by the inverse of that fraction to acheive the same result, let's do that: 

 

$$\left({\frac{\left(\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{z}}\right){\mathtt{\,\times\,}}\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}\right)\right)}{\left(\left({\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{z}}{\mathtt{\,-\,}}{\mathtt{x}}\right){\mathtt{\,\times\,}}\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right)\right)}}\right){\mathtt{\,\times\,}}\left({\frac{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}\right)$$

 

We can once again see a common term (4x - 3), let's cancel and combine fractions and re-write: 

$$\left({\frac{\left(\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{z}}\right){\mathtt{\,\times\,}}\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}\right)\right)}{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{z}}{\mathtt{\,-\,}}{\mathtt{x}}\right)}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}\right) = {\frac{\left(\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{z}}\right){\mathtt{\,\times\,}}\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}\right)\right)}{\left(\left({\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{z}}{\mathtt{\,-\,}}{\mathtt{x}}\right){\mathtt{\,\times\,}}\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)\right)}}$$

Now we have (x-3z)/(3z-x). This simplifies to -1. Let's cancel this final term and we are left with: 

$${\mathtt{\,-\,}}\left({\frac{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}}\right)$$

This is the final simplified answer.