It is given that $y = x^{xy}$, where $x > 0$, $y > 0$.

1. Translate 'tangent parallel to the y-axis' into a condition

A tangent parallel to the $y$-axis is vertical, so $\dfrac{dy}{dx}$ is undefined. Hence the denominator of $\dfrac{dy}{dx}=\dfrac{y^2(1+\ln x)}{1-xy\ln x}$ must be zero:
$1 - xy\ln x = 0 \quad\Rightarrow\quad xy\ln x = 1.$

2. Use the original curve relation to find y

Taking logarithms of $y = x^{xy}$ gives $\ln y = xy\ln x$. Substituting $xy\ln x = 1$:
$\ln y = 1 \quad\Rightarrow\quad y = e.$

3. Find the x-coordinate using a graphing calculator

Substitute $y = e$ into $xy\ln x = 1$:
$e\,x\ln x = 1 \quad\Rightarrow\quad x\ln x = \tfrac{1}{e}.$
Solving with a GC gives $x \approx 1.32$.
Hence the point is $(1.32,\ e) \approx (1.32,\ 2.72)$.