$\text { The rate of change of the volume of a sphere with respect to its surface area } \mathrm{S} \text { is }$

Let $r$ be the radius of the sphere. The volume $V$ and surface area $S$ are given by $V = \dfrac{4}{3} \pi r^3$ and $S = 4 \pi r^2$.

From the expression for surface area, $r^2 = \dfrac{S}{4 \pi}$, which implies $r = \sqrt{\dfrac{S}{4 \pi}} = \dfrac{1}{2} \sqrt{\dfrac{S}{\pi}}$.

Differentiating $V$ with respect to $r$, we get $\dfrac{dV}{dr} = 4 \pi r^2$.

Differentiating $S$ with respect to $r$, we get $\dfrac{dS}{dr} = 8 \pi r$.

The rate of change of volume with respect to surface area is $\dfrac{dV}{dS} = \dfrac{dV/dr}{dS/dr} = \dfrac{4 \pi r^2}{8 \pi r} = \dfrac{r}{2}$.

Substituting the value of $r$ in terms of $S$, we get $\dfrac{dV}{dS} = \dfrac{1}{2} \left( \dfrac{1}{2} \sqrt{\dfrac{S}{\pi}} \right) = \dfrac{1}{4} \sqrt{\dfrac{S}{\pi}}$.

Answer: $\dfrac{1}{4} \sqrt{\dfrac{S}{\pi}}$