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 $a-b=$

Given $af\left(x\right)+bf\left(\frac{1}{x}\right)=x+1 ...\left(i\right)$

Replacing $x\to \frac{1}{x}$, we get

$af\left(\frac{1}{x}\right)+bf\left(x\right)=\frac{1}{x}+1 ...\left(ii\right)$

$a\times \left(i\right)-b\times \left(ii\right)$, we get

$\left[a^{2}f\left(x\right)+abf\left(\frac{1}{x}\right)\right]-\left[abf\left(\frac{1}{x}\right)+b^{2}f\left(x\right)\right]=ax+a-\frac{b}{x}-b$

$\left(a^2-b^2\right)f\left(x\right)=ax-\frac{b}{x}+a-b$

$f\left(x\right)=\frac{ax-\frac{b}{x}}{a^2-b^2}+\frac{1}{a+b}$

Now, $\frac{d}{dx}\left(x^{2}f\left(x\right)\right)=\frac{{\displaystyle d}}{{\displaystyle dx}}\left(x^2\left(\frac{ax-\frac{b}{x}}{a^2-b^2}+\frac{1}{a+b}\right)\right)$

$=\frac{{\displaystyle d}}{{\displaystyle dx}}\left(\left(\frac{a{x}^3-bx}{a^2-b^2}+\frac{x^2}{a+b}\right)\right)=\frac{3a{x}^2-b}{a^2-b^2}+\frac{2x}{a+b}$

Given, $\frac{3a{x}^2}{a^2-b^2}+\frac{2x}{a+b}-\frac{b}{a^2-b^2}=2x^2+2x+\frac{1}{3}$

i.e. $\frac{3a}{a^2-b^2}=2$,

$\frac{1}{a+b}=1$ and 

$-\frac{b}{a^2-b^2}=\frac{1}{3}$

So, $\frac{a}{a-b}=\frac{{\displaystyle 2 }}{3}\& -\frac{b}{a-b}=\frac{1}{3}$

i.e. $a=-2b\Rightarrow b=-1, a=2$

Hence, $a-b=3$