If $\frac14 < x^2 < 4$, then we can write
$$\begin{align}
\lim_{n\to \infty}\left(4^n+x^{2n}+x^{-2n}\right)^{1/n}&=\lim_{n\to \infty}4\left(1+\left(\frac{x^2}{4}\right)^n+\left(\frac{x^{-2}}{4}\right)^n\right)^{1/n}\\\\
&=4
\end{align}$$
If $x^2>4$, then
$$\begin{align}
\lim_{n\to \infty}\left(4^n+x^{2n}+x^{-2n}\right)^{1/n}&=\lim_{n\to \infty}x^2\left(1+\left(\frac{4}{x^2}\right)^n+\left(\frac{1}{x^4}\right)^n\right)^{1/n}\\\\
&=x^2
\end{align}$$
If If $x^2<1/4$, then
$$\begin{align}
\lim_{n\to \infty}\left(4^n+x^{2n}+x^{-2n}\right)^{1/n}&=\lim_{n\to \infty}x^{-2}\left(1+\left(4x^2\right)^n+\left(x^4\right)^n\right)^{1/n}\\\\
&=x^{-2}
\end{align}$$
Therefore, we can write
$$f(x)=\begin{cases}x^2&,|x|>2\\\\4&\frac12\le |x|\le 2\\\\\frac{1}{x^2}&,0<|x|<\frac12\end{cases}$$
Obviously, $f$ is differentiable everywhere it is defined except when $x\ne \pm \frac12$ and $x=\pm 2$. Note that $f$ is undefined at $x=0$.