If $f(x) = \frac{\sin^{-1} x}{\sqrt{1- x ^2}}$, then evaluate $(1-x^2)f''(x) - xf(x)$ 
$f(x) = \dfrac{\sin^{-1} x}{\sqrt{1- x ^2}}$

Differentiating the given function, we get
$f'(x) = \dfrac{1 + \dfrac{x\sin^{-1}x}{\sqrt{1-x^2}}}{1-x^2}$
which can also be written as 

$f'(x) = \dfrac{1 + xf(x)}{1-x^2}$

Differentiating this one more time
$f''(x) = \dfrac{(1-x^2)[xf'(x) + f(x)] - [1 + xf(x)](-2x)}{(1-x^2)^2}$
Now the next obvious step is to substitute for $f'(x)$ since we have no $f'(x)$ term in the question.
Which further simplify the equation
$f''(x) = \dfrac{3x[1 + xf(x)] + (1-x^2)f(x)}{(1-x^2)^2}$
Head scratch, I have no idea how to proceed. Possibilities are I have made a mistake above, if so please point out and help me take this ahead.

Your question has probably a typo, here's how. 
Given $$f(x)=\frac{\sin^{-1}x}{\sqrt{1-x^2}}$$
Using quotient rule, 
$$f'(x)=\frac{1+\frac{x\sin^{-1}x}{\sqrt{1-x^2}}}{1-x^2}$$
Re arranging this we get, 
$$(1-x^2)f'(x)=1+x\times \frac{\sin^{-1}x}{\sqrt{1-x^2}}$$
$$(1-x^2)f'(x)=1+xf(x)$$
We get finally, $$(1-x^2)f\color{red}{'}(x)-xf(x)=\color{blue}{1}$$
This is the statement, which gives you the same result as mentioned in comments.
