Answer
$$y'=\sinh^{-1}({\frac{x}{3}})$$
Work Step by Step
$y'=\frac{d}{dx}(x\sinh^{-1}(\frac{x}{3})-\sqrt{9+x^2})$
$=\frac{d}{dx}x\sinh^{-1}(\frac{x}{3})-\frac{d}{dx}\sqrt{9+x^2}$
Using the product rule:
$y'=(1\times\sinh^{-1}(\frac{x}{3})+x\times\frac{d}{dx}\sinh^{-1}(\frac{x}{3})-\frac{d}{dx}\sqrt{9+x^2}$
Using the chain rule:
$y'=(\sinh^{-1}(\frac{x}{3})+x\times\frac{d\sinh^{-1}(\frac{x}{3})}{d\frac{x}{3}}\times
\frac{d\frac{x}{3}}{dx})-\frac{d\sqrt{9+x^2}}{d9+x^2}
\times\frac{d9+x^2}{dx}$
$=\sinh^{-1}({\frac{x}{3}})+x\times\frac{1}{\sqrt{1+(\frac{x}{3})^2}}
\times\frac{1}{3}
-\frac{1}{2\sqrt{9+x^2}} \times2x$
$=\sinh^{-1}({\frac{x}{3}})+\frac{x}{3\sqrt{\frac{x^2+9}{9}}}-\frac{x}{\sqrt{9+x^2}}$
$=\sinh^{-1}({\frac{x}{3}})+\frac{x}{3\times\frac{1}{3}\sqrt{{x^2+9}}}-\frac{x}{\sqrt{x^2+9}}$
$=\sinh^{-1}({\frac{x}{3}})+\frac{x}{\sqrt{{x^2+9}}}-\frac{x}{\sqrt{x^2+9}}$
$=\sinh^{-1}({\frac{x}{3}})$
