Quotient rules and fractions! I'm doing a problem in my textbook that I have the answer too thanks to it being an odd problem in the back. The problem is that I can't get to the answer of this problem.

Problem by the way is shown to be "Differentiate: (1 / y^2 - 3 / y^4)(y + 5y^3)".

The answer in the book happens to be: "5 + 14/ y^2 + 9/Y^4".

What I've tried thus far is use (fg)' = fg' + gf'

With F being the first parenthesis and G being the second parentheses respectably.

But doing so gets me (y^2 - 3)(5y^2 + 1) / y^3 + (y^2 - 3)(5y^2 + 1). I stop at this point because I know theirs no way to get the answer from here. 

Can someone point me to the right way?Also how can I make fractions like others are doing on this site? I can't find the button on here..

$\frac{1}{y^2}$
You can expand what you have to give:
$$\left(\frac{1}{y^2}-\frac{3}{y^4}\right)\left(y+5y^3\right)=5y-14y^{-1}-3y^{-3}$$
Therefore, the derivative will be given by:
$$\frac{d}{dy} (5y-14y^{-1}-3y^{-3})=5\cdot\frac{d}{dy}(y)-14\cdot\frac{d}{dy}(y^{-1})-3\cdot\frac{d}{dy}(y^{-3})$$
Now, use the fact that $\frac{d}{dy}(y^n)=ny^{n-1}$.



Alternatively, you can use the product rule or the quotient rule, as you have mentioned. I personally prefer to expand it in this case.

I see that you asked how to format questions using LaTeX. Here is a nice guide showing how to use the formatting.
So, you are taking the derivative with respect to y. What you have to do is to first expand your expression by multiplying it out. You will get

$\frac{1+5y^2}{y}-\frac{3+15y^2}{y^3}$ after some simplification. 

Then you simply take the derivative using the quotient rule: $\frac{f'g-fg'}{g^2}$ and you'll have it!