一条曲线的方程为 $x^3+3x^2y-y^3=3$。

(a) 证明

(b) 求曲线上切线平行于 $x$ 轴的点的坐标。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

分别对三项求导，其中 $3x^2y$ 使用乘积法则。收集导数后约去共同常数 $3$。

Differentiating implicitly,

Thus

Dividing and simplifying,

水平切线要求分子为零。先分解为 $x(x+2y)=0$，两个因子都要处理，再检查分母。

For a horizontal tangent,

If $x=0$, the curve equation gives $-y^3=3$, so

If $x=-2y$, substitution gives

Thus $y=1$ and $x=-2$.

At $(0,-\sqrt[3]{3})$, the denominator $y^2-x^2$ is positive; at $(-2,1)$ it is $-3$. Both are non-zero. Hence the required points are