已知一条曲线的方程为 $y = x^3 - 3x^2 + 3x - 7$。

1. Find the first derivative of the curve equation

$\frac{dy}{dx} = 3x^2 - 6x + 3$

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

For stationary points, $\frac{dy}{dx} = 0$.
$3x^2 - 6x + 3 = 0$
$x^2 - 2x + 1 = 0$
$(x - 1)^2 = 0$
$x = 1$

3. Find the corresponding $y$-coordinate

When $x = 1$,
$y = (1)^3 - 3(1)^2 + 3(1) - 7$
$y = 1 - 3 + 3 - 7 = -6$
The coordinates of the stationary point are $(1, -6)$.

1. Recall the derivative from part (a) and test the sign of the gradient for values of $x$ just before and after the stationary point at $x = 1$.

From part (a), $\frac{dy}{dx} = 3x^2 - 6x + 3 = 3(x - 1)^2$.

Let us test values of $x$ around $x = 1$:
When $x = 0.9$, $\frac{dy}{dx} = 3(0.9 - 1)^2 = 3(-0.1)^2 = 0.03 > 0$ (Positive)
When $x = 1$, $\frac{dy}{dx} = 0$
When $x = 1.1$, $\frac{dy}{dx} = 3(1.1 - 1)^2 = 3(0.1)^2 = 0.03 > 0$ (Positive)

2. Determine the nature of the stationary point based on the gradient signs.

Since the gradient is positive on both sides of the stationary point (it does not change sign), the point $(1, -6)$ is a stationary point of inflexion.