Step 1: Build the same constraint $r^2=h(2R-h)$, but organise the maximisation using $h/R$ as a single ratio variable to keep numbers clean:
Let $h=kR$ for some $0<k<2$. Then $r^2=kR(2R-kR)=R^2k(2-k)$, and $V=\dfrac13\pi r^2h=\dfrac13\pi R^2k(2-k)\cdot kR=\dfrac{\pi R^3}{3}k^2(2-k)$.
Step 2: Maximise the function $g(k)=k^2(2-k)=2k^2-k^3$ over $k$ instead of $h$ directly:
$g'(k)=4k-3k^2=k(4-3k)$. Setting $g'(k)=0$ gives $k=0$ (rejected, no cone) or $k=4/3$.
Step 3: Confirm it is a maximum using the second derivative:
$g''(k)=4-6k$; at $k=4/3$, $g''=4-8=-4<0$, confirming a maximum.
Step 4: Compute the maximum volume using this $k$:
$g(4/3)=(4/3)^2(2-4/3)=\dfrac{16}{9}\cdot\dfrac{2}{3}=\dfrac{32}{27}$, so $V_{\max}=\dfrac{\pi R^3}{3}\cdot\dfrac{32}{27}=\dfrac{32\pi R^3}{81}$, matching the direct-$h$ computation.
Step 5: Take the ratio to the sphere volume:
$\dfrac{V_{\max}}{(4/3)\pi R^3}=\dfrac{32\pi R^3/81}{4\pi R^3/3}=\dfrac{32}{81}\cdot\dfrac{3}{4}=\dfrac{8}{27}$.
Final Answer:
The largest inscribed cone's volume is $\dfrac{8}{27}$ of the sphere's volume.
\[ \boxed{\dfrac{8}{27}V_{\text{sphere}}} \]