If 8f(x)+6f($\frac{1}{x}$)=x+5 and y = x²f(x) then $\frac{\mathrm{dy}}{\mathrm{dx}}$ at x=-1

given 8f(x)+6f(1/x)=x+5

Replace x by $\frac{1}{x}$
$8\mathrm{f}(1/\mathrm{x})+6\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}+5$ 

By solving these two we get
$\begin{array}{l}\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{28}\left(8\mathrm{x}-\frac{6}{\mathrm{x}}+10\right)\\ \therefore \mathrm{y}={\mathrm{x}}^2\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{28}\left(8{\mathrm{x}}^3-6\mathrm{x}+10{\mathrm{x}}^2\right)\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{28}\left(24{\mathrm{x}}^2-6+20\mathrm{x}\right)\\ Now \frac{\mathrm{dy}}{\mathrm{dx}} at \mathrm{x}=-1=-\frac{1}{14}\end{array}$