If  $af\left(x\right)+bf\left(\frac{1}{x}\right)=x+1$  and  $\frac{d}{dx}\left(x^{2}f\left(x\right)\right)=2x^2+2x+\frac{1}{3}$  then a-b= 

 $\frac{d}{dx}\left(x^{2}f\left(x\right)\right)=2x^2+2x+\frac{1}{3}$
Integrating on both sides
 $$\begin{gathered} x^{2}f\left(x\right)=\frac{2x^3}{3}+\frac{2x^2}{2}+\frac{1}{3}x \\ f\left(x\right)=\frac{2}{3}x+1+\frac{1}{3x} \\ f\left(\frac{1}{x}\right)=\frac{2}{3x}+1+\frac{x}{3} \\ \therefore af\left(x\right)+bf\left(\frac{1}{x}\right)=x+1 \\ \Rightarrow (\frac{2a}{3}+\frac{b}{3})x+(\frac{a}{3}+\frac{2b}{3})\frac{1}{x}+(a+b)=x+1 \\ \frac{2a}{3}+\frac{b}{3}=1,a+b=1 \\ \Rightarrow a=2,b=-1 \\ \therefore a-b=3 \end{gathered}$$