Let $f$ be $a$differentiable function such that $8f\left(x\right)+6f(1/x)-x=5(x\ne 0)$ and $y=x^{2}f\left(x\right)$ then 

$\frac{dy}{dx}$ at $x=1$ is

 Differentiating the given expression, we get

$\begin{array}{r}8f^{\mathrm{\prime }}\left(x\right)+6f^{\mathrm{\prime }}\left(\frac{1}{x}\right)\left(-\frac{1}{x^2}\right)-1=0\\ \Rightarrow 8f^{\mathrm{\prime }}\left(1\right)+6f^{\mathrm{\prime }}\left(1\right)(-1)=1\\ \Rightarrow f^{\mathrm{\prime }}\left(1\right)=1/2\end{array}$

Also $\frac{dy}{dx}=2xf\left(x\right)+x^2{f}^{\mathrm{\prime }}\left(x\right)$

so,  ${\left.\frac{dy}{dx}\right|}_{x=1}=2f\left(1\right)+f^{\mathrm{\prime }}\left(1\right)$

Putting $x = 1$ in the given equation, we obtain

$\begin{array}{l}\Rightarrow 8f\left(1\right)+6f\left(1\right)-1=5\\ 14f\left(1\right)=6\\ \Rightarrow f\left(1\right)=\frac{3}{7}\end{array}$

Hence      ${\left.\frac{dy}{dx}\right|}_{x=1}=2,\frac{3}{7}+\frac{1}{2}=\frac{19}{14}$