Answer
Replace $x$ by $x^2$ in the Maclaurin series for $f(x)$
Converses for $|x|<1$
Work Step by Step
We are given the Maclaurin series for $f$ that converges for $|x|<1$.
The Maclaurin series for $f(x)$ is $\sum_{k=0}^{\infty} \dfrac{f^{(k)}(0)}{k!}x^k$.
The Maclaurin series for $f(x^2)$ is obtained by replacing $x$ by $x^2$:
$\sum_{k=0}^{\infty} \dfrac{f^{(k)}(0)}{k!}x^{2k}$.
The Maclaurin series for $f(x^2)$ converges for:
$|x^2|<1\Rightarrow |x|<1$
