Use exp/log functions to rewrite y as
[tex]y = (1+x)^{\frac1x} \implies y = \exp\left(\ln\left(1+x)^{\frac1x}\right)\right)[/tex]
One of the properties of logarithms lets us bring down the exponent, so
[tex]y = \exp\left(\dfrac{\ln(1+x)}x\right)[/tex]
Now take the derivative of both sides with respect to x. By the chain and quotient rules,
[tex]\dfrac{dy}{dx} = \exp\left(\dfrac{\ln(1+x)}x\right) \times \dfrac{\frac x{1+x} - \ln(1+x)}{x^2}[/tex]
Simplify:
[tex]\boxed{\dfrac{dy}{dx} = \frac{x - (1+x)\ln(1+x)}{x^2(1+x)} (1+x)^{\frac1x}}[/tex]
