Answer
$$y'=\frac{(2x-\sin^{-1}y)\sqrt{1-y^2}}{x}$$
Work Step by Step
$$x\sin^{-1}y=1+x^2$$
Using implicit differentiation, we differentiate both sides of the equation with respect to $x$: $$x'\sin^{-1}y+x(\sin^{-1}y)'=0+2x$$
$$\sin^{-1}y+x\times\frac{1}{\sqrt{1-y^2}}\times y'=2x$$
$$\sin^{-1}y+\frac{x}{\sqrt{1-y^2}}\times y'=2x$$
Next, we need to separate elements with $y'$ and those without $y'$ into 2 sides of the equation:
$$\frac{x}{\sqrt{1-y^2}}\times y'=2x-\sin^{-1}y$$
Finally, calculate for $y'$: $$y'=\frac{2x-\sin^{-1}y}{\frac{x}{\sqrt{1-y^2}}}$$
$$y'=\frac{(2x-\sin^{-1}y)\sqrt{1-y^2}}{x}$$
