Solution verification: finding a Maclaurin series for $f$, interval of convergence, and $f^{(10)}(0)$ I have to find Maclaurin series for function $f(x)$ = $2x^2\over{16+x^4}$, it's interval of convergence and $f^{(10)}(0)$. I managed to calculate Maclaurin series and $10^{th}$ derivative, but I'm not sure if it's done in proper way and if solution is correct.

On determining the series,
$$\begin{align}
f(x) &= \frac{2x^2}{16+x^4} \\
&= \frac{2x^2}{16} \cdot \frac{1}{1-\frac{-x^4}{16}} \\
&= \frac{2x^2}{16} \cdot \sum_{i=0}^\infty \left(-\frac{x^4}{16} \right)^n  \\
&= \frac{2x^2}{16} \cdot \sum_{i=0}^\infty \frac{(-1)^n \cdot x^{4n}}{16^n} \\
&= \sum_{i=0}^\infty \frac{(-1)^n \cdot x^{4n+2}}{2^{4n+3}}
\end{align}$$
for $\vert{-x^4\over16}\vert<1 \implies x\in(-2;2)$

On determining $f^{(10)}(0)$,
$$f^{(10)}(0)\cdot \frac{ x^{10}}{ 10!} = \frac{x^{42}}{2^{43}} \implies f^{(10)}(0) = x^{32}\cdot \frac{10!}{2^{43}}$$

Looking for feedback and opinion if the way I solved it is correct.

Your computation of the MacLaurin series of $f$ looks right, but remember that $f^{(10)}(0)$ is a number. You know that$$\frac{f^{(10)}(0)}{10!}=\frac{(-1)^2}{2^{11}}.$$Therefore,$$f^{(10)}(0)=\frac{10!}{2^{11}}.$$
Also, $\left\lvert\frac{-x^4}{16}\right\rvert<1$ doesn't just imply that $x\in(-2,2)$; it is actually equivalent to it.
