If you know the Taylor expansion of the elementary function and the rules to compose one Taylor expansion into another, you can write the Taylor expansion of every function which is expressed by the composition of elementary functions, without computing any derivative.
This can be made a lot simpler by changing variables. (Changing variables is commonly taught as a technique for integration, but it can also be handy for differentiation.)
Introduce the new variable $u=1-x$. Then $x=1-u$, and
$$f(x) = \frac{1-u}{\sqrt{u}} = u^{-1/2} - u^{1/2}$$
If we define a new function $g(x)=x^{-1/2} - x^{1/2}$ then this tells us that $$f(x) = g(1-x),$$ and therefore on taking derivatives we have $$f^{(n)}(x) = (-1)^n g^{(n)}(1-x)$$
This change of variables allows you to essentially swap out the problem of computing derivatives of $f(x)$ and trade it for computing derivatives of the (much simpler) function $g(x)$.
Now, the derivatives of $g(x)$ are
$$g'(x) = \left( - \frac{1}{2}\right)x^{-3/2} - \left(\frac{1}{2}\right)x^{-1/2} $$
$$g''(x) = \left( - \frac{1}{2}\right)\left( - \frac{3}{2}\right)x^{-5/2} - \left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)x^{-3/2} $$
$$g'''(x) = \left( - \frac{1}{2}\right)\left( - \frac{3}{2}\right)\left( - \frac{5}{2}\right)x^{-7/2} - \left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)x^{-5/2} $$
and in general if we introduce the notation $A_n$ to denote the product of the first $n$ odd numbers (for example, $A_1=1$, $A_2 = 1\cdot 3$, $A_3 = 1\cdot 3 \cdot 5$, etc.) then
$$g^{(n)}(x)=(-1)^n \frac{A_n}{2^n}x^{-(2n+1)/2} + (-1)^n \frac{A_{n-1}}{2^n}x^{-(2n-1)/2}$$
Now we recall that $f^{(n)}(x) = (-1)^n g^{(n)}(1-x)$, so that
$$f^{(n)}(x)=\frac{A_n}{2^n}(1-x)^{-(2n+1)/2} + \frac{A_{n-1}}{2^n}(1-x)^{-(2n-1)/2}$$
The only thing left is to express the coefficients $A_n$ in a more convenient closed form; for that, see Proving formula for product of first n odd numbers.
Best Answer
$$2f(x)=\frac1{1+ix}+\frac1{1-ix}.$$ Therefore $$2f^{(n)}(x)=\frac{(-i)^nn!}{(1+ix)^{n+1}}+\frac{i^nn!}{(1-ix)^{n+1}}.$$