Let f : R + → R + be a function satisfying f x - x = λ (constant), ∀ x ∈ R + and f x…

Given relation is $f\left(x . f\left(y\right)\right)=f\left(xy\right)+x .....\left(1\right)$

Interchanging $x$ and y in equation $\left(1\right)$ we get,

$f\left(y . f\left(x\right)\right)=f\left(yx\right)+y .......\left(2\right)$

Again replacing $x$ with $f\left(x\right)$ in equation $\left(2\right)$ we get

$f\left(f\left(x\right)·f\left(y\right)\right)=f\left(y·f\left(x\right)\right)+f\left(x\right)$

$\Rightarrow f\left(f\left(x\right)·f\left(y\right)\right)=f\left(yx\right)+y+f\left(x\right) ...\left(3\right)$ {from equation $2$}

Again interchanging $x$ and $y$ in equation $\left(3\right)$, we have

$f\left(f\left(y\right)·f\left(x\right)\right)=f\left(yx\right)+x+f\left(y\right).......\left(4\right)$

Now from equation $\left(3\right) \& \left(4\right)$ we get,

$f\left(xy\right)+y+f\left(x\right)=f\left(yx\right)+x+f\left(y\right) ......\left(5\right)$

Suppose $f\left(x\right)-x=f\left(y\right)-y=\lambda$

Substituting $f\left(x\right)=\lambda +x$ in equation $\left(5\right)$ we have

$x·f\left(y\right)+\lambda =\left(xy+\lambda \right)+x$

$\Rightarrow x·f\left(y\right)=xy+x$

Therefore $x\left(y+\lambda \right)=xy+x \left[\because f\left(y\right)=\lambda +y\right]$

$\Rightarrow \lambda =x\Rightarrow \lambda =1 \left(\because x>0\right)$

So $f\left(x\right)=x+\lambda =x+1$

Hence $\underset{x\to 0}{\lim }\frac{{\left(f\left(x\right)\right)}^{{\displaystyle \frac{1}{3}}}}{{\left(f\left(x\right)\right)}^{{\displaystyle \frac{1}{2}}}-1}=\underset{x\to 0}{\lim }\frac{{\left(1+x\right)}^{{\displaystyle \frac{1}{3}}}-1}{{\left(1+x\right)}^{{\displaystyle \frac{1}{2}}}-1}$

$=\underset{x\to 0}{\lim }\left(\frac{{\left(1+x\right)}^{{\displaystyle \frac{1}{3}}}-1}{1+x-1}\right)·\left(\frac{1+x-1}{{\left(1+x\right)}^{{\displaystyle \frac{1}{2}}}-1}\right)$$=\frac{{\displaystyle \frac{1}{3}}}{{\displaystyle \frac{1}{2}}}=\frac{2}{3}$