Solution:

Let’s take $^{y}=\left(x+\frac{1}{x}\right)^{x}+x^{\left(x-\frac{1}{x}\right)}$

Putting the value of $\left(x+\frac{1}{x}\right)^{x}=u$ and $x^{\left(x+\frac{1}{x}\right)}=v$

$y=u+v$

$\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} \ldots \dots \dots$(1)

So now $u=\left(x+\frac{1}{x}\right)^{x}$

$\log u=\log \left(x+\frac{1}{x}\right)^{x}=x \log \left(x+\frac{1}{x}\right)$

$\frac{1}{u} \frac{d u}{d x}=x \cdot \frac{1}{\left(x+\frac{1}{x}\right)} \frac{d}{d x}\left(x+\frac{1}{x}\right)+\log \left(x+\frac{1}{x}\right) \cdot 1$

$\frac{1}{u} \frac{d u}{d x}=x \cdot \frac{1}{\left(x+\frac{1}{x}\right)}\left(x-\frac{1}{x^{2}}\right)+\log \left(x+\frac{1}{x}\right) \cdot 1$

$\frac{d u}{d x}=u\left[\frac{x^{2}-1}{x^{2}+1}+\log \left(x+\frac{1}{x}\right)\right]$

$=\left(x+\frac{1}{x}\right)^{x}\left[\frac{x^{2}-1}{x^{2}+1}+\log \left(x+\frac{1}{x}\right)\right] . \ldots \dots \dots$(2)

Now again $v=x^{\left(x+\frac{1}{x}\right)}$

$\log v=\log x^{\left(x+\frac{1}{x}\right)}=\left(x+\frac{1}{x}\right) \log x$

$\frac{1}{v} \frac{d v}{d x}=\left(x+\frac{1}{x}\right) \cdot \frac{1}{x}+\log x\left(\frac{-1}{x^{2}}\right)$

$\frac{d v}{d x}=v\left[\frac{1}{x}\left(x+\frac{1}{x}\right)-\frac{1}{x^{2}} \log x\right]$

$\frac{d v}{d x}=x^{\left(x+\frac{1}{x}\right)}\left[\frac{1}{x}\left(x+\frac{1}{x}\right)-\frac{1}{x^{2}} \log x\right] \ldots \dots \dots$(3)

Putting the values from eq.(2) and eq.(3) in eq.(1),

$\frac{d y}{d x}=\left(x+\frac{1}{x}\right)^{x}\left[\frac{x^{2}-1}{x^{2}+1}+\log \left(x+\frac{1}{x}\right)\right]+x^{\left(x+\frac{1}{x}\right)}\left[\frac{1}{x}\left(x+\frac{1}{x}\right)-\frac{1}{x^{2}} \log x\right]$