If \(8 \mathrm{f}(x)+6 \mathrm{f}\left(\frac{1}{x}\right)=x+5\) and \(y=x^2 \mathrm{f}(x)\), then \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) at \(x=-1\) is

\(8 f(x)+6 f\left(\frac{1}{x}\right)=x+5...(i)\)
Differentiating w.r.t. \(x\), we get
\(8 f^{\prime}(x)+6 f^{\prime}\left(\frac{1}{x}\right)\left(\frac{-1}{x^2}\right)=1...(ii)\)
Substituting \(x=-1\) in (i), we get \(8 f(-1)+6 f(-1)=4\)
\(\therefore 14 f(-1)=4 \)
\( \therefore f(-1)=\frac{4}{14}=\frac{2}{7}...(iii)\)
Substituting \(x=-1\) in (ii), we get
\(\begin{aligned}
& \mathrm{f}^{\prime}(-1)=\frac{1}{2} \\.
& y=x^2 \mathrm{f}(x)
\end{aligned}\)...(iv)
\(\therefore \frac{\mathrm{d} y}{\mathrm{~d} x} =2 \)
\( \left.\therefore \frac{\mathrm{~d} y}{\mathrm{~d} x}\right|_{x=-1} =2(-1)+x^2 \mathrm{f}^{\prime}(x) \)
\( =\frac{-4}{7}+\frac{1}{2} \ldots[\text { From (iii) and (iv) }] \)
\( =-\frac{1}{14}\)