Answer
$$r'=\frac{\sec\sqrt\theta\tan\sqrt\theta\tan\frac{1}{\theta}}{2\sqrt\theta}-\frac{\sec\sqrt\theta\sec^2\frac{1}{\theta}}{\theta^2}$$
Work Step by Step
$$r=\sec\sqrt\theta\tan\frac{1}{\theta}$$
The derivative of function $r$ is $$r'=(\sec\sqrt\theta)'\tan\frac{1}{\theta}+\sec\sqrt\theta\Big(\tan\frac{1}{\theta}\Big)'$$
According to the Chain Rule, we have
$$(\sec\sqrt\theta)'=\sec\sqrt\theta\tan\sqrt\theta(\sqrt\theta)'=\sec\sqrt\theta\tan\sqrt\theta(\theta^{1/2})'$$ $$=\frac{1}{2}\sec\sqrt\theta\tan\sqrt\theta(\theta^{-1/2})=\frac{\sec\sqrt\theta\tan\sqrt\theta}{2\sqrt\theta}$$
$$\Big(\tan\frac{1}{\theta}\Big)'=\sec^2\frac{1}{\theta}\Big(\frac{1}{\theta}\Big)'=\sec^2\frac{1}{\theta}\Big(\frac{-1(\theta)'}{\theta^2}\Big)=\sec^2\frac{1}{\theta}\Big(-\frac{1}{\theta^2}\Big)$$ $$=-\frac{1}{\theta^2}\sec^2\frac{1}{\theta}$$
Substitute these results back to the calculation of $r'$: $$r'=\frac{\sec\sqrt\theta\tan\sqrt\theta\tan\frac{1}{\theta}}{2\sqrt\theta}-\frac{\sec\sqrt\theta\sec^2\frac{1}{\theta}}{\theta^2}$$
