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

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

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

2. Expand the right-hand side and make $\frac{\text{d}y}{\text{d}x}$ the subject.

$1 + \frac{\text{d}y}{\text{d}x} = 2x - 2y - (2x - 2y)\frac{\text{d}y}{\text{d}x}$
$\frac{\text{d}y}{\text{d}x} + (2x - 2y)\frac{\text{d}y}{\text{d}x} = 2x - 2y - 1$
$\frac{\text{d}y}{\text{d}x}(1 + 2x - 2y) = 2x - 2y - 1$
$\frac{\text{d}y}{\text{d}x} = \frac{2x - 2y - 1}{1 + 2x - 2y}$

3. Evaluate $1 - \frac{\text{d}y}{\text{d}x}$ by substituting the derived expression and simplifying to a single fraction.

$1 - \frac{\text{d}y}{\text{d}x} = 1 - \frac{2x - 2y - 1}{1 + 2x - 2y}$
$= \frac{1 + 2x - 2y - (2x - 2y - 1)}{1 + 2x - 2y}$
$= \frac{1 + 2x - 2y - 2x + 2y + 1}{1 + 2x - 2y}$
$= \frac{2}{1 + 2x - 2y}$

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

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

2. Factor the expression and substitute the result from part (a) to prove the given identity.

$-\frac{\text{d}^2 y}{\text{d}x^2} = -4(1 + 2x - 2y)^{-2} \left(1 - \frac{\text{d}y}{\text{d}x}\right)$
$\frac{\text{d}^2 y}{\text{d}x^2} = \left(\frac{2}{1 + 2x - 2y}\right)^2 \left(1 - \frac{\text{d}y}{\text{d}x}\right)$
From part (a), we know that $\frac{2}{1 + 2x - 2y} = 1 - \frac{\text{d}y}{\text{d}x}$. Substituting this gives:
$\frac{\text{d}^2 y}{\text{d}x^2} = \left(1 - \frac{\text{d}y}{\text{d}x}\right)^2 \left(1 - \frac{\text{d}y}{\text{d}x}\right)$
$\frac{\text{d}^2 y}{\text{d}x^2} = \left(1 - \frac{\text{d}y}{\text{d}x}\right)^3$

1. Evaluate the second derivative at the turning point.

At a turning point, the first derivative is zero, so $\frac{\text{d}y}{\text{d}x} = 0$.
Substituting this into the expression for the second derivative from (b):
$\frac{\text{d}^2 y}{\text{d}x^2} = (1 - 0)^3 = 1$

2. Determine the nature of the turning point.

Since $\frac{\text{d}^2 y}{\text{d}x^2} = 1 > 0$, by the second derivative test, the turning point is a minimum.