An straightforward differentiation

By chain rule, $\dfrac{\mathrm d}{\mathrm dx}\left(x+\dfrac 1x\right)^3=3\left(x+\dfrac 1x\right)^2\left(1-\dfrac 1{x^2}\right)$
Evaluated at $x=1$ , it is $3\left(1+\dfrac 11\right)^2\left(1-\dfrac 1{1^2}\right)=0$

$x+ \frac{1}{x^{3}}=x^{3}+ \frac{1}{x^{3}}+ \frac{3}{x^{2}}+3x^{2}$ so $\frac{\mathrm{d} (x + \frac{1}{x})}{\mathrm{d} x} = \frac{\mathrm{d} (x^{3}+\frac{1}{x^{3}}+\frac{3}{x}+3x}{\mathrm {d}x}$ = $3x^{2}-\frac{3}{x^{2}}+3-\frac{3}{x^4}$ $\rightarrow$ so at x =1 this would sum up to 3-3+3-3=0 . And dont mind the solution the 1st one is better in just practising latex .

Can you tell me more about the d/dx