If the Taylor series for $e^x$ is known, the Taylor Series for $e^{x^2}$ can be found by replacing all $x$ with $x^2$.
What about a Taylor Series known for the following $f(x)=\ln\frac{1+x}{1-x}=2\displaystyle\sum\limits_{n=0}^ \infty \frac{x^{2n+1}}{2n+1}$. Can the Taylor Series of $f(x)=\ln\frac{1+x^2}{1-x^2}$ be found by replacing all $x$ with $x^2$ - making it $2\sum^\infty_{n=0}\frac{x^{4n+2}}{2n+1}$?
Yes, this method would work. Because $x$ was replaced with $x^2$ in the original function, you just make that change in the series as well.