To complete your computation, let's treat the case of a function supported in interval $(0,1)$. Indeed, for $ f\in C^\infty_c(0,1)$ there is an inequality $$ \int_0^1 r^{-3}f(r)^6 dr\le C\left(\int_0^1 rf''(r)^2dr \right)^3\\ .$$

For any $ f\in C^\infty _ c(\mathbb{R} _ + )$, the Hardy inequality with exponent $3$, applied to the function $f(r)^2$, gives $$\int_0^\infty r^{-3}f(r)^6 dr=\int_0^\infty\left( \frac{f(r)^2}{r}\right)^3 dr\le 3^3\int_0^{\infty }f(r)^3f'(r)^3 dr\\ .$$ By the Cauchy-Schwarz inequality $$\left( \int_0^{\infty }f(r)^3f'(r)^3 dr \right)^2\le \left( \int_0^{\infty }r^{-3}f(r)^6 dr \right) \left( \int_0^{\infty }r^3f'(r)^6 dr \right) $$ so that $$\int_0^\infty r^{-3}f(r)^6 dr\le 3^6\int_0^{\infty }r^3f'(r)^6 dr \\ .\qquad\qquad(1)$$ Now assume $\operatorname{supp}(f)\subset(0,1)$, and let $u\in C^{\infty}(\mathbb{R} _ +)$ be the composition $u(p):=f'(e^{-p})$. With the change of variable $p:=-\log r$, the latter integral writes $$\int_0^1 r^3f'(r)^6 dr= \int_0^{\infty } e^{-4p}u(p)^6 dp\\ .\qquad\qquad (2)$$ Since $u(0)=0$, we have the well-known pointwise bounds on $u$ $$|u(p)|\le \int_0^p|u'(p)| dp\le p^{1/2}\left(\int_0^{\infty}|u'(p)|^2 dp \right)^{1/2}$$ so, by (2), returning to the $r$ variables $$\int_0^1 r^3f'(r)^6 dr\le\left(\int_0^{\infty} p^3 e^{-4p} dp \right) \left(\int_0^{\infty}|u'(p)|^2 dp \right)^3 =\frac{3}{2^7}\left(\int_0^1 rf''(r)^2dr \right)^3\\ .$$ The latter, together with inequality (1) proves the claim with $C:=(3/2)^7\\ .$

To complete your computation, let's treat the case of a function supported in interval $(0,1)$. Indeed, for $ f\in C^\infty_c(0,1)$ there is an inequality $$ \int_0^1 r^{-3}f(r)^6 dr\le C\left(\int_0^1 rf''(r)^2dr \right)^3\, .$$

For any $ f\in C^\infty _ c(\mathbb{R} _ + )$, the Hardy inequality with exponent $3$, applied to the function $f(r)^2$, gives $$\int_0^\infty r^{-3}f(r)^6 dr=\int_0^\infty\left( \frac{f(r)^2}{r}\right)^3 dr\le 3^3\int_0^{\infty }f(r)^3f'(r)^3 dr\, .$$ By the Cauchy-Schwarz inequality $$\left( \int_0^{\infty }f(r)^3f'(r)^3 dr \right)^2\le \left( \int_0^{\infty }r^{-3}f(r)^6 dr \right) \left( \int_0^{\infty }r^3f'(r)^6 dr \right) $$ so that $$\int_0^\infty r^{-3}f(r)^6 dr\le 3^6\int_0^{\infty }r^3f'(r)^6 dr \, .\qquad\qquad(1)$$ Now assume $\operatorname{supp}(f)\subset(0,1)$, and let $u\in C^{\infty}(\mathbb{R} _ +)$ be the composition $u(p):=f'(e^{-p})$. With the change of variable $p:=-\log r$, the latter integral writes $$\int_0^1 r^3f'(r)^6 dr= \int_0^{\infty } e^{-4p}u(p)^6 dp\, .\qquad\qquad (2)$$ Since $u(0)=0$, we have the well-known pointwise bounds on $u$ $$|u(p)|\le \int_0^p|u'(p)| dp\le p^{1/2}\left(\int_0^{\infty}|u'(p)|^2 dp \right)^{1/2}$$ so, by (2), returning to the $r$ variables $$\int_0^1 r^3f'(r)^6 dr\le\left(\int_0^{\infty} p^3 e^{-4p} dp \right) \left(\int_0^{\infty}|u'(p)|^2 dp \right)^3 =\frac{3}{2^7}\left(\int_0^1 rf''(r)^2dr \right)^3\, .$$ The latter, together with inequality (1) proves the claim with $C:=(3/2)^7\, .$