It is given that $y=x^{y}$, where $x>0$, $y>0$. Find $\dfrac{dy}{dx}$ in terms of x and y. Hence find the equation of the tangent to the curve $y=x^{y}$ which is parallel to the y-axis.

1. Take the natural logarithm of both sides to prepare for implicit differentiation.

$y = x^y$
$\ln y = \ln (x^y)$
$\ln y = y \ln x$

2. Differentiate implicitly with respect to $x$.

$\frac{d}{dx}(\ln y) = \frac{d}{dx}(y \ln x)$
Using the product rule on the right hand side:
$\frac{1}{y} \frac{dy}{dx} = \frac{dy}{dx} \ln x + y \left( \frac{1}{x} \right)$

3. Rearrange the equation to make $\frac{dy}{dx}$ the subject.

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

4. Identify the condition for the tangent to be parallel to the y-axis.

A tangent parallel to the $y$-axis has an undefined gradient (it is a vertical line), which occurs when the denominator of $\frac{dy}{dx}$ is zero.
$x(1 - y \ln x) = 0$
Since we are given $x > 0$:
$1 - y \ln x = 0 \implies y \ln x = 1$

5. Solve for $y$ using the initial logarithmic equation.

From Step 1, we have the identity $\ln y = y \ln x$.
Substitute $y \ln x = 1$ into this equation:
$\ln y = 1$
$y = e$

6. Solve for $x$ and state the equation of the tangent line.

Substitute $y = e$ back into $y \ln x = 1$:
$e \ln x = 1$
$\ln x = \frac{1}{e}$
$x = e^{\frac{1}{e}}$
Since the tangent is a vertical line, its equation is simply $x$ equals the x-coordinate of this point:
$x = e^{\frac{1}{e}}$