Answer
$$y'=\tanh^{-1}x$$
Work Step by Step
$y'=\frac{d}{dx}(x\tanh^{-1}x+\ln{\sqrt{1-x^2}})$
$=\frac{d}{dx}x\tanh^{-1}x+\frac{d}{dx}\ln\sqrt{1-x^2}$
Using the product rule:
$y'=(\tanh^{-1}x+x\frac{d}{dx}\tanh^{-1}x)+\frac{d}{dx}\ln{\sqrt{1-x^2}}$
$=\tanh^{-1}x+x\times\frac{1}{1-x^2}+\frac{d}{dx}\ln{\sqrt{1-x^2}}$
$=\tanh^{-1}x+\frac{x}{1-x^2}+\frac{d}{dx}\ln{\sqrt{1-x^2}}$
Using the chain rule:
$y'=\tanh^{-1}x+\frac{x}{1-x^2}+\frac{d\ln{\sqrt{1-x^2}}}{d{\sqrt{1-x^2}}}
\times\frac{d{\sqrt{1-x^2}}}{d1-x^2}
\times\frac{d1-x^2}{dx}$
$=\tanh^{-1}x+\frac{x}{1-x^2}+\frac{1}{\sqrt{1-x^2}}
\times\frac{1}{2\sqrt{1-x^2}}
\times -2x$
$=\tanh^{-1}x+\frac{x}{1-x^2}+\frac{-x}{{1-x^2}}$
$=\tanh^{-1}x$
