<s>[INST] 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.
 [/INST] we will differentiate using the implicit definition of $y.$  we have $$y = \frac{\sin^{-1} x}{\sqrt {1- x^2}} \to y^2(1-x^2) = (\sin^{-1} x)^2 \tag 1$$ differentiating once, we have $$ -2xy^2 +2yy'(1-x^2) =\frac 2{\sqrt{1-x^2}}\sin^{-1} x = 2y$$  diving out by $y, $ you get $$1+xy  =y'(1-x^2) \tag 2 $$ differentiating $(2),$ we have $$xy'  + y = -2xy' +y''(1-x^2) \to (1-x^2)y''-x'y =y+2xy'=y+\frac{2x(1+xy)}{1-x^2}\\=\frac{y(1+x^2)+2x}{1-x^2} $$
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