The equation of a curve is $y = e^{x^2 - 5x}$.

1. Differentiate $y$ with respect to $x$ using the chain rule

$\frac{dy}{dx} = e^{x^2 - 5x} \cdot \frac{d}{dx}(x^2 - 5x)$
$\frac{dy}{dx} = (2x - 5)e^{x^2 - 5x}$

2. Differentiate $\frac{dy}{dx}$ with respect to $x$ using the product rule

Let $u = 2x - 5$ and $v = e^{x^2 - 5x}$.
$\frac{du}{dx} = 2$
$\frac{dv}{dx} = (2x - 5)e^{x^2 - 5x}$
$\frac{d^2y}{dx^2} = u \frac{dv}{dx} + v \frac{du}{dx}$
$\frac{d^2y}{dx^2} = (2x - 5)(2x - 5)e^{x^2 - 5x} + e^{x^2 - 5x}(2)$

3. Simplify the expression for the second derivative

$\frac{d^2y}{dx^2} = [(2x - 5)^2 + 2]e^{x^2 - 5x}$
$\frac{d^2y}{dx^2} = (4x^2 - 20x + 25 + 2)e^{x^2 - 5x}$
$\frac{d^2y}{dx^2} = (4x^2 - 20x + 27)e^{x^2 - 5x}$

1. Set the first derivative to zero to find the $x$-coordinate of the stationary point

From part (i), $\frac{dy}{dx} = (2x - 5)e^{x^2 - 5x}$.
For a stationary point, $\frac{dy}{dx} = 0$.
$(2x - 5)e^{x^2 - 5x} = 0$

2. Solve for $x$

Since $e^{x^2 - 5x} > 0$ for all real $x$, we have:
$2x - 5 = 0$
$x = \frac{5}{2}$

3. Find the corresponding $y$-coordinate

Substitute $x = \frac{5}{2}$ into the equation of the curve:
$y = e^{(\frac{5}{2})^2 - 5(\frac{5}{2})}$
$y = e^{\frac{25}{4} - \frac{25}{2}}$
$y = e^{-\frac{25}{4}}$

4. State the exact coordinates of the stationary point

The coordinates of the stationary point are $(\frac{5}{2}, e^{-\frac{25}{4}})$.