Let k and m be positive real number such that the function $\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}3{\mathrm{x}}^2+\mathrm{k}\sqrt{\mathrm{x}+1},\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0<\mathrm{x}<1\\ {\mathrm{mx}}^2+{\mathrm{k}}^2 ,\hspace{0.25em} \mathrm{x}\ge 1\end{array}\right.$
is differentiable for all x > 0. Then $\frac{8{\mathrm{f}}^{\mathrm{\prime }}\left(8\right)}{{\displaystyle {\mathrm{f}}^{\mathrm{\prime }}\left(\frac{1}{8}\right)}}$ is equal to

$\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}3{\mathrm{x}}^2+\mathrm{k}\sqrt{\mathrm{x}+1}\\ 3{\mathrm{mx}}^2+{\mathrm{k}}^2\end{array}\right.$
$\text{ For continuity at }x=1$
$\begin{array}{l}f\left(1^{-}\right)=f\left(1^{+}\right)\\ \Rightarrow 3+k\sqrt{2}=m+k^2\end{array}$
$\begin{array}{l}{\left.f^{\mathrm{\prime }}\left(x\right)\right|}_{x=1^{-}}=6x+{\left.\frac{k}{2\sqrt{x+1}}\right|}_{x=1^{-}}=6+\frac{k}{2\sqrt{2}}\\ {\left.f^{\mathrm{\prime }}\left(x\right)\right|}_{x=1^{+}}={\left.2mx\right|}_{x=1^{+}}=2m\end{array}$
$\text{ For differentiability at }x=1$
$6+\frac{k}{2\sqrt{2}}=2m$
$k=\frac{7}{4\sqrt{2}}\text{ and }m=\frac{103}{32}$
$\begin{array}{l}{f}^{\mathrm{\prime }}\left(8\right)={\left.2mx\right|}_{x=8}=16m=\frac{103}{2}\\ f^{\mathrm{\prime }}\left(\frac{1}{8}\right)={\left.\left(6x+\frac{k}{2\sqrt{x+1}}\right)\right|}_{x=\frac{1}{8}}=\frac{4}{3}\\ \frac{8\cdot f^{\mathrm{\prime }}\left(8\right)}{f^{\mathrm{\prime }}\left(\frac{1}{8}\right)}=309\end{array}$