Given: \(y = x^{\frac{1}{x}}\)

Taking the natural logarithm of both sides:

\(\ln y = \ln \left(x^{\frac{1}{x}}\right)\)

\(\ln y = \frac{1}{x} \ln x\)

Differentiating both sides with respect to \(x\):

\(\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \left(\frac{1}{x} \ln x\right)\)

Using the product rule:

\(\frac{1}{y} \frac{dy}{dx} = \left(-\frac{1}{x^2}\right) \ln x + \frac{1}{x} \left(\frac{1}{x}\right)\)

\(\frac{1}{y} \frac{dy}{dx} = -\frac{\ln x}{x^2} + \frac{1}{x^2}\)

\(\frac{1}{y} \frac{dy}{dx} = \frac{1 - \ln x}{x^2}\)

Now, solve for \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = y \left(\frac{1 - \ln x}{x^2}\right)\)

\(\frac{dy}{dx} = x^{\frac{1}{x}} \left(\frac{1 - \ln x}{x^2}\right)\)

Evaluate \(\frac{dy}{dx}\) at \(x = 1\):

\(\left.\frac{dy}{dx}\right|_{x=1} = (1)^{\frac{1}{1}} \left(\frac{1 - \ln 1}{1^2}\right)\)

\(\left.\frac{dy}{dx}\right|_{x=1} = 1 \left(\frac{1 - 0}{1}\right)\)

\(\left.\frac{dy}{dx}\right|_{x=1} = 1\)