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I was going through some problems in high school textbook and stumbled on this problem.
I could be trying to solve it whole day and I wouldn't solve it. The solution seems to be too complicated for high school student (I marked the part red I can't understand). Can you explain how do you go about solving this problem and what is going through your mind when doing so?

One problem comprehending such arguments is inessential information is obfuscating the algebra, e.g. the fractional exponents. They have a common denominator of $3$ so writing $,x = a^{1/3}, y = b^{1/3}$ simplifies it, making the steps much clearer, namely

$$dfrac{-2x^3y^4 + 2 x^5y^2}{x^6y^4 - y^6 x^4} = dfrac{2x^3y^2 (color{#c00}{-y^2+x^2})}{x^4 y^4, ( color{#c00}{x^2,-,y^2})} = dfrac{2}{xy^2} = dfrac{2}{a^{1/3}b^{2/3}} $$

Through practice, you become more familiar with various techniques which you can use to make simplification easier and faster.

The previous answer provides a clever substitution which simplifies the problem greatly, but you can solve the question rather easily even without it. In the example, you have

$$frac{-2ab^{frac{4}{3}}+2a^{frac{5}{3}}b^{frac{2}{3}}}{a^2b^{frac{4}{3}}-b^2a^{frac{4}{3}}}$$

In order to understand the steps given for the solution, you can break the problem down into two parts. Focusing on the numerator first, you can rewrite it as

$$-2abb^{frac{1}{3}}+2aa^{frac{2}{3}}b^{frac{2}{3}}$$

because we have $a^{b+c} = a^ba^c$, so, for instance $b^{frac{4}{3}} = b^{1+frac{1}{3}} = bb^{frac{1}{3}}$. Using this technique again, you can split $b$ into $b^{frac{1}{3}}b^{frac{2}{3}}$, so you get

$$-color{blue}{2ab^{frac{2}{3}}}b^{frac{1}{3}}b^{frac{1}{3}}+color{blue}{2a}a^{frac{2}{3}}color{blue}{b^{frac{2}{3}}}$$

Notice how both sides share common factors. You have $ab+ac = a(b+c)$, so the expression can once more be rewritten as

$$2ab^{frac{2}{3}}left(-b^{frac{2}{3}}+a^{frac{2}{3}}right) tag{1}$$

In fact, in the picture you’ve given, the simplification wasn’t immediately made. Rather, all the steps were shown, as in $ab+ac = aleft(frac{ab}{a}+frac{ac}{a}right)$ rather than immediately $ab+ac = a(b+c)$.

Repeat the same process for the denominator (rewriting and factoring):

$$a^2b^{frac{4}{3}}-b^2a^{frac{4}{3}} = a^{frac{2}{3}}color{blue}{a^{frac{4}{3}}b^{frac{4}{3}}}-b^{frac{2}{3}}color{blue}{b^{frac{4}{3}}a^{frac{4}{3}}} = a^{frac{4}{3}}b^{frac{4}{3}}left(a^{frac{2}{3}}-b^{frac{2}{3}}right) tag{2}$$

Putting it all together, you reach

$$frac{2ab^{frac{2}{3}}color{blue}{left(-b^{frac{2}{3}}+a^{frac{2}{3}}right)}}{a^{frac{4}{3}}b^{frac{4}{3}}color{blue}{left(a^{frac{2}{3}}-b^{frac{2}{3}}right)}} = frac{2ab^{frac{2}{3}}}{a^{frac{4}{3}}b^{frac{4}{3}}} = frac{2ab^{frac{2}{3}}}{aa^{frac{1}{2}}b^{frac{2}{3}}b^{frac{2}{3}}} = frac{2}{a^{frac{1}{3}}b^{frac{2}{3}}}$$

As you can see, there really was nothing apart from manipulating the exponents and factoring. At first, this might seem long, tricky, and time-consuming. You need some practice in order to see these ideas and patterns quickly, and you’ll eventually solve these problems very quickly.