Passage: Let $\left(e^y\right)^x-y=0$, where $y$ is a function of $x$ whose domain is $(0,10]$.

Question: What is $\dfrac{dy}{dx}$ equal to, given that $y=y_0$ when $x=1$ ?

Given the equation $(e^y)^x - y = 0$, we can rewrite it as:

$e^{xy} = y$

Taking the natural logarithm on both sides:

$xy = \ln y$

Differentiating both sides with respect to $x$ using the product rule and chain rule:

$y + x \dfrac{dy}{dx} = \dfrac{1}{y} \dfrac{dy}{dx}$

Rearranging the terms to solve for $\dfrac{dy}{dx}$:

$y = \left( \dfrac{1}{y} - x \right) \dfrac{dy}{dx}$

$y = \left( \dfrac{1 - xy}{y} \right) \dfrac{dy}{dx}$

$\dfrac{dy}{dx} = \dfrac{y^2}{1 - xy}$

We are given that at $x = 1$, $y = y_0$. Substituting these values into the derivative:

$\dfrac{dy}{dx} = \dfrac{y_0^2}{1 - y_0}$

From the original equation $e^{xy} = y$, substituting $x = 1$ and $y = y_0$ gives:

$e^{y_0} = y_0$

To match the given options, we substitute $y_0 = e^{y_0}$ for one $y_0$ in the numerator and the $y_0$ in the denominator:

$\dfrac{dy}{dx} = \dfrac{y_0 e^{y_0}}{1 - e^{y_0}}$

Answer: $\dfrac{y_0 e^{y_0}}{1-e^{y_0}}$