曲线 C 的方程为 $x + y = (x - y)^2$。 已知 C 只有一个转向点。

1. Differentiate both sides of 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 rearrange the equation to make $\frac{dy}{dx}$ the subject.

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

3. Substitute the expression for $\frac{dy}{dx}$ into $1 - \frac{dy}{dx}$ and simplify to obtain the required result.

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

1. Differentiate the result from part (a) implicitly with respect to $x$.

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

Differentiating both sides with respect to $x$ using the chain rule:
$-\frac{d^2y}{dx^2} = -2(1 + 2x - 2y)^{-2} \cdot \frac{d}{dx}(1 + 2x - 2y)$
$-\frac{d^2y}{dx^2} = -2(1 + 2x - 2y)^{-2} \left(2 - 2\frac{dy}{dx}\right)$

2. Simplify the differentiated equation.

Factor out $2$ from the bracket on the right hand side:
$-\frac{d^2y}{dx^2} = -4(1 + 2x - 2y)^{-2} \left(1 - \frac{dy}{dx}\right)$

Divide by $-1$ to isolate $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{4}{(1 + 2x - 2y)^2} \left(1 - \frac{dy}{dx}\right)$

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

Notice that $\frac{4}{(1 + 2x - 2y)^2}$ is the square of $\frac{2}{1 + 2x - 2y}$.

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

Substitute this identity into the second derivative expression:
$\frac{d^2y}{dx^2} = \left(1 - \frac{dy}{dx}\right)^2 \left(1 - \frac{dy}{dx}\right)$
$\frac{d^2y}{dx^2} = \left(1 - \frac{dy}{dx}\right)^3$

1. Substitute the condition for a turning point into the expression for the second derivative.

At the turning point, $\frac{dy}{dx} = 0$.
Substitute this into the expression for $\frac{d^2 y}{dx^2}$ given in part (b):
$\frac{d^2 y}{dx^2} = \left(1 - 0\right)^3 = 1$

2. Use the second derivative test to conclude the nature of the turning point.

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