This is a problem that I have not been able to solve. I don't know how to approach differential equations when used in intervals ...
Find $\dfrac{d^2y}{dx^2}$, if $$1-\left(\frac{dy}{dx}\right)^2 \geq \sec^2 p$$ for $p\in \operatorname{dom}(\sec x)$.
Following William Elliot's comment above, given $p\in\mathbb{R}$ fixed and abbreviating $\frac{dy}{dx}=y'$ we have $$1-\sec^2 p = 1- \frac{1}{\cos^2(p)} \geq y'^2.$$ Because $0 \leq \cos(p)^2 \leq 1$, we have $\frac{1}{\cos^2(p)}=\sec^ 2 p \geq 1$ and thus $$1-\sec^2 p\leq 0.$$ As $y'^2\geq 0$, the original inequality implies $y'(x)=0$ (for whatever values of $x$ the original inequality is supposed to hold). In particular, if $y'(x)=0$ for all $x$, we find $$\frac{d^2y}{dx^2}(x)= \frac{d}{dx}y'(x)=0$$ for all $x$.