曲线 C 的方程为 $x + y = (x - y)^2.$ 已知 C 只有一个驻点。

1. Differentiate the given equation implicitly with respect to $x$.

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

2. Expand and make $\frac{dy}{dx}$ the subject of the equation.

$1 + \frac{dy}{dx} = 2(x - y) - 2(x - y)\frac{dy}{dx}$
$\frac{dy}{dx} + 2(x - y)\frac{dy}{dx} = 2(x - y) - 1$
$\frac{dy}{dx}(1 + 2x - 2y) = 2x - 2y - 1$
$\frac{dy}{dx} = \frac{2x - 2y - 1}{1 + 2x - 2y}$

3. Evaluate and simplify $1 - \frac{dy}{dx}$ to match the required form.

$1 - \frac{dy}{dx} = 1 - \frac{2x - 2y - 1}{1 + 2x - 2y}$
$1 - \frac{dy}{dx} = \frac{(1 + 2x - 2y) - (2x - 2y - 1)}{1 + 2x - 2y}$
$1 - \frac{dy}{dx} = \frac{1 + 2x - 2y - 2x + 2y + 1}{1 + 2x - 2y}$
$1 - \frac{dy}{dx} = \frac{2}{1 + 2x - 2y} \quad \text{(shown)}$

1. Differentiate both sides of the equation from part (a) with respect to $x$.

From part (a), we have:
$1 - \dfrac{dy}{dx} = 2(1 + 2x - 2y)^{-1}$

Differentiating both sides with respect to $x$:
$-\dfrac{d^2y}{dx^2} = -2(1 + 2x - 2y)^{-2} \cdot \dfrac{d}{dx}(1 + 2x - 2y)$

2. Apply the chain rule to complete the differentiation.

Evaluating the derivative of the inner function:
$-\dfrac{d^2y}{dx^2} = -2(1 + 2x - 2y)^{-2} \left(2 - 2\dfrac{dy}{dx}\right)$

Factorising out $2$:
$-\dfrac{d^2y}{dx^2} = -4(1 + 2x - 2y)^{-2} \left(1 - \dfrac{dy}{dx}\right)$

Multiplying by $-1$ and rewriting with a positive exponent:
$\dfrac{d^2y}{dx^2} = \dfrac{4}{(1 + 2x - 2y)^2} \left(1 - \dfrac{dy}{dx}\right)$

3. Substitute the expression from part (a) to show the final result.

From part (a), we know that $\dfrac{2}{1 + 2x - 2y} = 1 - \dfrac{dy}{dx}$. Squaring both sides gives:
$\dfrac{4}{(1 + 2x - 2y)^2} = \left(1 - \dfrac{dy}{dx}\right)^2$

Substituting this into the equation for the second derivative:
$\dfrac{d^2y}{dx^2} = \left(1 - \dfrac{dy}{dx}\right)^2 \left(1 - \dfrac{dy}{dx}\right)$
$\dfrac{d^2y}{dx^2} = \left(1 - \dfrac{dy}{dx}\right)^3$ (Shown)

1. State the condition for a turning point and evaluate the second derivative using the given identity.

At a turning point, $\frac{dy}{dx} = 0$.

Substituting $\frac{dy}{dx} = 0$ into the expression from (b):
$\frac{d^2y}{dx^2} = (1 - 0)^3$
$\frac{d^2y}{dx^2} = 1$

2. Conclude the nature of the turning point using the second derivative test.

Since $\frac{d^2y}{dx^2} = 1 > 0$, the turning point is a minimum.