Answer
$$ y'=\frac{\sec \theta (\theta \tan \theta -1) }{\theta ^2}$$
Work Step by Step
Recall that $(\sec x)'=\sec x\tan x$.
Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$
Since $ y= \frac{\sec \theta }{\theta }$, then, using the quotient rule, we have \begin{align*}
y'&=\frac{\theta (\sec \theta)'- \sec \theta (\theta )'}{\theta ^2}\\
&=\frac{\theta \sec \theta \tan \theta -\sec \theta }{\theta ^2}\\
&=\frac{\sec \theta (\theta \tan \theta -1) }{\theta ^2}.
\end{align*}
