This is the differential equation $$\frac{d^2u}{d\theta^2}+u=-\frac{m}{l^2}\frac{d}{du}V\Big(\frac{1}{u}\Big)\tag 1$$

take $u(\theta)\mapsto u(-\theta)$

$$\frac{du}{d\theta}\,\mapsto-\frac{du}{d\theta}~~~, \frac{d^2u}{d\theta^2}\mapsto \frac{d^2u}{d\theta^2}$$ thus equation (1) $~\mapsto$ $$\frac{d^2u}{d\theta^2}+u=-\frac{m}{l^2}\frac{d}{du}V\Big(\frac{1}{u}\Big)\tag 2$$

equation (2) is equal equation (1)

Edit

$$y=u(f(\theta))\\ \frac{d y}{d\theta}=\frac{d}{d\theta}\,u(f(\theta))\frac{d f}{d\theta}$$

with $~f(\theta)=-\theta~$

$$\frac{d y}{d\theta}=\frac{d}{d\theta}\,u(-\theta)\,(-1)$$

This is the differential equation $$\frac{d^2u}{d\theta^2}+u=-\frac{m}{l^2}\frac{d}{du}V\Big(\frac{1}{u}\Big)\tag 1$$

take $u(\theta)\mapsto u(-\theta)$

$$\frac{du}{d\theta}\,\mapsto-\frac{du}{d\theta}~~~, \frac{d^2u}{d\theta^2}\mapsto \frac{d^2u}{d\theta^2}$$ thus equation (1) $~\mapsto$ $$\frac{d^2u(-\theta)}{d\theta^2}+u(-\theta)=-\frac{m}{l^2}\frac{d}{du}V\Big(\frac{1}{u(-\theta)}\Big)\tag 2$$

equation (2) is equal equation (1)

Edit

$$y=u(f(\theta))\\ \frac{d y}{d\theta}=\frac{d}{d\theta}\,u(f(\theta))\frac{d f}{d\theta}$$

with $~f(\theta)=-\theta~$

$$\frac{d y}{d\theta}=\frac{d}{d\theta}\,u(-\theta)\,(-1)$$

This is the differential equation $$\frac{d^2u}{d\theta^2}+u=-\frac{m}{l^2}\frac{d}{du}V\Big(\frac{1}{u}\Big)\tag 1$$

take $u(\theta)\mapsto u(-\theta)$

$$\frac{du}{d\theta}\,\mapsto-\frac{du}{d\theta}~~~, \frac{d^2u}{d\theta^2}\mapsto \frac{d^2u}{d\theta^2}$$ thus equation (1) $~\mapsto$ $$\frac{d^2u}{d\theta^2}+u=-\frac{m}{l^2}\frac{d}{du}V\Big(\frac{1}{u}\Big)\tag 2$$

equation (2) is equal equation (1)

This is the differential equation $$\frac{d^2u}{d\theta^2}+u=-\frac{m}{l^2}\frac{d}{du}V\Big(\frac{1}{u}\Big)\tag 1$$

take $u(\theta)\mapsto u(-\theta)$

$$\frac{du}{d\theta}\,\mapsto-\frac{du}{d\theta}~~~, \frac{d^2u}{d\theta^2}\mapsto \frac{d^2u}{d\theta^2}$$ thus equation (1) $~\mapsto$ $$\frac{d^2u}{d\theta^2}+u=-\frac{m}{l^2}\frac{d}{du}V\Big(\frac{1}{u}\Big)\tag 2$$

equation (2) is equal equation (1)

Edit

$$y=u(f(\theta))\\ \frac{d y}{d\theta}=\frac{d}{d\theta}\,u(f(\theta))\frac{d f}{d\theta}$$

with $~f(\theta)=-\theta~$

$$\frac{d y}{d\theta}=\frac{d}{d\theta}\,u(-\theta)\,(-1)$$