You are looking at the intersection $C = C_1 \cap C_2$ of two cones, $C_1$ defined by the condition $f(x) \ge 0$ and $C_2$ defined by the condition $Df(x) \ge 0$, where $D=(1+x\partial_x)(1-x\partial_x)$. In a sense, $C_2 = D^{-1} C_1$, where the right-hand-side is interpreted as the pre-image of $C_1$ with respect to $D$.

Now, $C_1$ does not have extreme rays (those that do not belong to the sum of any two different rays from the same cone), because any non-negative function with non-trivial support can always be written as a convex linear combination with positive coefficients of two other non-negative functions with smaller supports. If you embed $C^2(\mathbb{R}_{++})$ in distributions, then then the closure of $C_1$ does have extreme rays, which are the Dirac $\delta$ distributions, $\delta(x-y)$.

Next, look at $C_2$. Using an explicit right-inverse $D^{-1}_0$ of the operator $D$, we can write $C_2 = \ker D + D^{-1}_0 C_1$, where $\ker D = \mathbb{R}_+ x + \mathbb{R}_+ x^{-1}$. A right inverse can be given, if I got all the calculations right, by the integral operator \begin{multline*} D^{-1}_0[g](x) = \int_0^x \frac{(x-y)(x+y)}{2xy^2} \chi_0(y) g(y) \, dy \\ + \int_x^\infty \frac{(y-x)(x+y)}{2xy^2} \chi_\infty(y) g(y) \, dy , \end{multline*} where $\chi_0(y)$ and $\chi_\infty(y)$ are any two non-negative continuous functions such that $\chi_0(y) + \chi_\infty(y) = 1$, with $\chi_0(y)$ zero on a neighborhood of $0$ and $\chi_\infty(y)$ zero on a neighborhood of infinity. The way it's written, the integrals are well-defined for any continuous $g(x)$. Note that, because the integral kernel is non-negative, $g(x) \ge 0$ implies that $D^{-1}_0[g](x) \ge 0$.

Finally, with the above observations, we can conclude that $$ C = C_1 \cap (\ker D + D^{-1}_0 C_1) = D^{-1}_0 C_1 + \ker_{\ge} D , $$ where $\ker_{\ge} D = \mathbb{R}_+ x + \mathbb{R}_+ x^{-1}$. Now, $D^{-1}C_1$ does not have any extreme rays either, by the same argument as for $C_1$. On the other hand, $\ker_{\ge} D$ does have extreme rays, those generated by $x$ and $x^{-1}$. Moreover, since $\ker_{\ge} D \cap D^{-1}_0 C_1 = \varnothing$, these two rays remain extreme when we take the sum of these two cones to get $C$.

So, it seems that the conclusion is that the cone $C$ has only the two extreme rays generated by $x$ and $x^{-1}$ and no other ones. Clearly, $C$ is very far from being even the closure of the convex linear combinations of its extreme rays.

You are looking at the intersection $C = C_1 \cap C_2$ of two cones, $C_1$ defined by the condition $f(x) \ge 0$ and $C_2$ defined by the condition $Df(x) \ge 0$, where $D=(1+x\partial_x)(1-x\partial_x)$. In a sense, $C_2 = D^{-1} C_1$, where the right-hand-side is interpreted as the pre-image of $C_1$ with respect to $D$.

Now, $C_1$ does not have extreme rays (those that do not belong to the sum of any two different rays from the same cone), because any non-negative function with non-trivial support can always be written as a convex linear combination with positive coefficients of two other non-negative functions with smaller supports. If you embed $C^2(\mathbb{R}_{++})$ in distributions, then then the closure of $C_1$ does have extreme rays, which are the Dirac $\delta$ distributions, $\delta(x-y)$.

Next, look at $C_2$. Using an explicit right-inverse $D^{-1}_0$ of the operator $D$, we can write $C_2 = \ker D + D^{-1}_0 C_1$, where $\ker D = \mathbb{R}_+ x + \mathbb{R}_+ x^{-1}$. A right inverse can be given, if I got all the calculations right (edit: turns out that I did miss an overall minus sign in the previous version), by the integral operator \begin{multline*} D^{-1}_0[g](x) = -\int_0^x \frac{(x-y)(x+y)}{2xy^2} \chi_0(y) g(y) \, dy \\ - \int_x^\infty \frac{(y-x)(x+y)}{2xy^2} \chi_\infty(y) g(y) \, dy , \end{multline*} where $\chi_0(y)$ and $\chi_\infty(y)$ are any two non-negative continuous functions such that $\chi_0(y) + \chi_\infty(y) = 1$, with $\chi_0(y)$ zero on a neighborhood of $0$ and $\chi_\infty(y)$ zero on a neighborhood of infinity. The way it's written, the integrals are well-defined for any continuous $g(x)$. Note that, because the integral kernel is non-positive, $g(x) \ge 0$ implies that $D^{-1}_0[g](x) \le 0$. Note that, because the integral kernel is non-negative, $g(x) \ge 0$ implies that $D^{-1}_0[g](x) \ge 0$.

So, it seems that the conclusion is that the cone $C$ has only the two extreme rays generated by $x$ and $x^{-1}$ and no other ones. Clearly, $C$ is very far from being even the closure of the convex linear combinations of its extreme rays.