Answer
Relative maximum: $x=\pm 2$
Relative minimum: $x=0$
Work Step by Step
First derivative
\[
f^{\prime}(x)=\frac{32 x-2 x^{5}}{\left(16+x^{4}\right)^{2}}=\frac{-2 x\left(x^{4}-16\right)}{\left(16+x^{4}\right)^{2}}
\]
Critical points (zeros of first derivative or point where the first derivative does not exist):
\[
x=\pm 2, x=0
\]
Second derivative
\[
f^{\prime \prime}(x)=\frac{-384 x^{4}+6 x^{8}+512}{\left(16+x^{4}\right)^{3}}
\]
Values of the second derivative at critical points:
\[
f^{\prime \prime}(-2)=-\frac{1}{8}<0
\]
\[
\begin{array}{l}
f^{\prime \prime}(0)=\frac{1}{8}>0 \\
f^{\prime \prime}(2)=-\frac{1}{8}<0
\end{array}
\]
Relative minimum: $x=0$
Relative maximum: $x=\pm 2$
