f(x) = xx has a stationary point at

The correct option is x = 1/e.

To find the stationary points of the function $f\left(x\right)=x^x$, we need to find the value of $x$ where its first derivative is equal to zero, i.e., $f\text{'}\left(x\right)=0$.

Let us define the function as:
$$y=x^x$$

To differentiate this variable-to-the-power-of-variable function, we take the natural logarithm ($\ln$) on both sides:
$$\ln \left(y\right)=\ln \left(x^x\right)$$

Using the logarithmic property $\ln \left(a^b\right)=b\ln \left(a\right)$, we can rewrite the equation as:
$$\ln \left(y\right)=x\ln \left(x\right)$$

Now, we differentiate both sides with respect to $x$. Applying the chain rule on the left side and the product rule on the right side, we get:
$$\frac{1}{y}\cdot \frac{dy}{dx}=\frac{d}{dx}\left[x\right]\cdot \ln \left(x\right)+x\cdot \frac{d}{dx}\left[\ln \right(x\left)\right]$$

Simplifying the derivatives:
$$\frac{1}{y}\cdot \frac{dy}{dx}=1\cdot \ln \left(x\right)+x\cdot \frac{1}{x}$$

This simplifies further to:
$$\frac{1}{y}\cdot \frac{dy}{dx}=\ln \left(x\right)+1$$

Multiplying both sides by $y$ allows us to isolate the derivative $\frac{dy}{dx}$:
$$\frac{dy}{dx}=y\cdot \left(\ln \right(x)+1)$$

Substituting back the original value of $y=x^x$, we obtain:
$$f\text{'}\left(x\right)=x^x\left(\ln \right(x)+1)$$

A stationary point occurs where the derivative is equal to zero:
$$x^x\left(\ln \right(x)+1)=0$$

Since $x^x>0$ for all valid real values of $x>0$ (the domain of the function), the only way for the product to be zero is if the term inside the parentheses is zero:
$$\ln \left(x\right)+1=0$$

Solving for $\ln \left(x\right)$:
$$\ln \left(x\right)=-1$$

Converting the logarithmic form to its exponential equivalent (with base $e$):
$$x=e^{-1}$$

Which is equivalent to:
$$x=\frac{1}{e}$$

Thus, the function has a stationary point at $x=\frac{1}{e}$.