Answer
$$ y'=\tan^{-1} x+\frac{x}{1+x^2}.$$
Work Step by Step
Recall the product rule: $(uv)'=u'v+uv'$
Recall that $(\tan^{-1} x)'=\dfrac{1}{1+x^2}$
Since $ y=x\tan^{-1} x $, the derivative $ y'$ by using the product rule, is given by
$$ y'=\tan^{-1} x+\frac{x}{1+x^2}.$$
