Taylor/Maclaurin Series Show that if x is small compared with unity, then $$f(x)=\frac{(1-x)^\frac{-2}{3}+(1-4x)^\frac{-1}{3}}{(1-3x)^\frac{-1}{3}+(1-4x)^\frac{-1}{4}}=1-\frac{7x^2}{36}$$
In my first attempt I expanded all four brackets up to second order of x, but this didn't lead me to something that could be expressed as the final result. In my second attempt I decided to find $f'(x)$ and $f''(x)$ and use these to find $f'(0)$ and $f''(0)$ to find the Maclaurin expansion of $f(x)$ but this was way too time consuming. Can someone lead me to right track and offer some assistance? Thank you 
 A: Expand each piece in the numerator and denominator up to the second order in Taylor series (this is the minimum taking into account what you are supposed to find). Then develop the result again as a Taylor series ... and you will find it !
A: For two real functions $f,g$, we have
$$
\left( \frac fg \right)' = \frac{f'g -fg'}{g^2}, \quad \left( \frac fg \right)'' = \frac{f''g -fg''}{g^2} - 2 \frac{f'g-fg'}{g^3}g'.
$$
It follows that the Taylor expansion of your function to the second order depends only on the Taylor expansion of your numerator and denominator to the second order. Let $f(x)$ be your numerator and $g(x)$ your denominator. Compute $f(0), f'(0), f''(0), g(0), g'(0), g''(0)$ and plug them in the last equations ; they will give you your Taylor expansion for $f/g$ around $x=0$, thus giving you the polynomial on the right.
For instance,
$$
f(x) = (1-x)^{-2/3} + (1-4x)^{-1/3} \\
f'(x) = \frac 23 (1-x)^{-5/3} + \frac 43 (1-4x)^{-4/3} \\
f''(x) = \frac{10}9 (1-x)^{-8/3} + \frac{64}{9}(1-4x)^{-7/3}
$$
hence $f(0) = 2$, $f'(0) = 2$ and $f''(0) = 74/9$.  Do the same for $g$, you'll get there!
Hope that helps,
A: The development of the last fraction (ratio of the two quadratic polynomials) is
1 - 7 x^2 / 36 + 7 x^3 / 36 + 35 x^4 / 144.
