If $f$ is uniformly continuous on $\mathbb{R}$, $f(x) \ge a >0$ and $g(x) = 1/f(x)^2$, then $g(x)$ is uniformly continuous I get to the this point 
$$
|g(x)-g(y)| = 
\left|\frac{1}{f(x)^2} - \frac{1}{f(y)^2}\right| = 
\left|\frac{f(x)^2 - f(y)^2}{f(x)^2f(y)^2}\right| \leq \frac{1}{a^4} |f(x)-f(y)|\,|f(x)+f(y)|.
$$
I want to use my assumption $|x-y| < \delta$, but I don't know how to do that given $|f(x)-f(y)|$.

An upper bound that is not as resourceful but also serves is as follows:
\begin{align*}
|g(x)-g(y)| 
=& \left|\frac{1}{f(x)^2} - \frac{1}{f(y)^2}\right|
\\
=& 
\left|\frac{f(x)^2 - f(y)^2}{f(x)^2f(y)^2}\right| 
\\
=&
\left|\frac{1}{f(x)^2} \right|\cdot \left|\frac{1}{f(y)^2} 
\right|\cdot \left|\color{red}{f(x)\cdot f(x)} - f(y)\cdot f(y)\right|
\\
= 
& 
\left|\frac{1}{f(x)^2} \right|\cdot \left|\frac{1}{f(y)^2} \right|
\cdot 
\left|\Big[ \color{red}{\big(f(x)-f(y)\big)}+ \color{blue}{f(y)} \Big]
\cdot \Big[ \color{red}{\big(f(x)-f(y)\big)}+ \color{blue}{f(y) }\Big] - f(y)\cdot f(y)\right|
\\
=
& 
\left|\frac{1}{f(x)^2} \right|\cdot \left|\frac{1}{f(y)^2} \right|
\cdot 
\bigg|
\color{red}{\Big(f(x)-f(y)\Big)\cdot \Big(f(x)-f(y)\Big)}
+2\cdot\color{blue}{f(y)}\cdot \color{red}{\Big(f(x)-f(y)\Big)}
\bigg|
\end{align*}
Due to do the triangle inequality, $\cfrac{1}{f(x)}\leq \cfrac{1}{a}$ and$ \cfrac{1}{f(y)}\leq \cfrac{1}{a}$:
\begin{align*}
|g(x)-g(y)|
\leq
& 
\left|\frac{1}{a^4} \right|
\cdot 
\Big|\color{red}{\big(f(x)-f(y)\big)}\Big|\cdot
\Big|\color{red}{\big(f(x)-f(y)\big)}\Big|
\\
&
\qquad\qquad +
\left|\frac{1}{a^3} \right|
\cdot 
\left|\frac{1}{f(y)} \right|
\cdot2\cdot\big|\color{blue}{f(y)}\big|\cdot \Big|\color{red}{\big(f(x)-f(y)\big)}\Big|
\\
\leq
& 
\left|\frac{1}{a^4} \right|
\cdot 
\Big|\color{red}{\big(f(x)-f(y)\big)}\Big|\cdot \Big|\color{red}{\big(f(x)-f(y)\big)}\Big|
+\left|\frac{2}{a^3} \right|
\cdot 
\Big|\color{red}{\big(f(x)-f(y)\big)}\Big|
\\
\end{align*}
