Answer
$$ y'= \frac{e^x+2x}{1-(e^x+x^2)^2} .$$
Work Step by Step
Since $ y= \tanh^{-1} (e^x+x^2)$, then the derivative, by using the chain rule and the fact that $(\tanh^{-1}x)'=\frac{dx}{1-x^2}$, is given by
$$ y'= \frac{1}{1-(e^x+x^2)^2} (e^x+x^2)'= \frac{e^x+2x}{1-(e^x+x^2)^2} .$$
