If $G(x)=e^{f(x)}$, by the chain rule:

${\displaystyle G\text{'}(x)=e^{f(x)}*f\text{'}(x).}$

As $e$ is never zero, the gradient function of $G(x)$ is zero when the gradient function of $f(x)$ is zero and it always has the same sign as $f\text{'}(x)$ so $G(x)$ and $f(x)$ have the same turning points.

To answer the question of turning points of $G(x)=x^{\frac{1}{x}}=e^{1/x\log x}$ let's first find the derivative of $\frac{1}{x}\log x$.

${\displaystyle \frac{d}{\mathrm{dx}}f(x)=\frac{d}{\mathrm{dx}}(\frac{1}{x}\log x)=\frac{1-\log x}{x^2}.}$

The derivative is positive for $\log x<1$, that is $x<e$, it is is zero when $\log x=1$, $x=e$ and it is negative for $x>e$. Hence the function is increasing for $x<e$ and decreasing for $x>e$.

${\displaystyle \frac{d^2}{{\mathrm{dx}}^2}(\frac{1}{x}\log x)=\frac{-x-2x(1-\log x)}{x^4}=\frac{2\log x-3}{x^3}.}$

At $x=e$ the second derivative of $f(x)$ is $-1/e^3$ which is negative so $f(x)$ has a maximum and so $G(x)$ has a maximum. Therefore $(e,e^{1/e})$ is a maximum point of $G(x)=x^{\frac{1}{x}}$.

To prove $lim{x}^{\frac{1}{x}}\to 1$ as $x\to \infty$ we may use the similar result for the discrete case $n^{\frac{1}{n}}$. As the graph of $f(x)=x^{\frac{1}{x}}$ is decreasing as $x\to \infty$ for each value of $x$ there is a value of $n$ for which $n^{1/n}<x^{1/x}$ and vice versa so both tend to the same limit which is 1.

What if $x\to 0$? Well, as suggested in the question, letting $1/x=t$ in $x^{\frac{1}{x}}$, one can note that as $x\to 0$, $t\to \infty$. Writing $t=1/x$,

${\displaystyle x^{1/x}={\left(\frac{1}{t}\right)}^t=\frac{1}{t^t}.}$

As $x\to 0$ we have $t\to \infty$ and $t^t\to \infty$ so

${\displaystyle x^{1/x}=\frac{1}{t^t}\to 0}$

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Hence the graph of $y=x^{1/x}$ is always above the line $y=0$ although the function is undefined at $x=0$.