Answer
$x=\displaystyle \pm\frac{2\sqrt{7}}{7}$
Work Step by Step
The slope of the tangent at x=a
is the derivative of f(x), calculated in x=a.
The tangent line has slope $f^{\prime}(a)$, and passes through $(a,f(a)).$
\begin{align*}
f(x)&=\displaystyle \frac{x}{(x^{2}+4)^{4}} \quad & \\
& \text{ quotient rule, } \left[\frac{u(x)}{v(x)}\right]^{\prime} =\displaystyle \frac{v(x)\cdot u^{\prime}(x)-u(x)\cdot v^{\prime}(x)}{[v(x)]^{2}}. & \\\\
u(x)&=x\quad u^{\prime}(x)=1 & \\
v(x)& =(x^{2}+4)^{4}\quad & \\
& \text{... apply chain rule } & \\
v^{\prime}(x)& =4(x^{2}+4)^{3}\cdot(x^{2}+4)^{\prime}\quad & \\
& =4(x^{2}+4)^{3}(2x)\quad & \\\\
f^{\prime}(x)& =\displaystyle \frac{(x^{2}+4)^{4}\cdot 1-x\cdot 4(x^{2}+4)^{3}(2x)}{(x^{2}+4)^{8}}\quad & \\
& =\displaystyle \frac{(x^{2}+4)^{3}[(x^{2}+4)-8x^{2}]}{(x^{2}+4)^{8}}\quad & \\
& =\displaystyle \frac{4-7x^{2}}{(x^{2}+4)^{5}}\quad & \\
\end{align*}
The derivative is zero when
\begin{align*}
4-7x^{2}&=0 \quad & \\
4&=7x^{2} \quad & \\
x^{2}&=\displaystyle \frac{4}{7} \quad & \\
x&=\displaystyle \pm\frac{2}{\sqrt{7}}\cdot\frac{\sqrt{7}}{7}\\
x&=\displaystyle \pm\frac{2\sqrt{7}}{7}
\end{align*}
The tangent line is horizontal at $x=\displaystyle \pm\frac{2\sqrt{7}}{7}$
