一条曲线的方程为 $y\mathrm{e}^{2x}+y^2\mathrm{e}^x=6$。

求曲线上 $y=1$ 处的斜率。

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。

先由 $y=1$ 解出正的 $e^x$，再做隐式求导并代入该点。

When $y=1$,

Let $u=e^x>0$. Then $u^2+u-6=(u+3)(u-2)=0$, so $u=2$ and $x=\ln2$.

Differentiating the original equation implicitly,

At $y=1$, $e^x=2$, let the gradient be $m$. Then

Thus $8m=-10$, giving $\boxed{m=-\frac54}$.

先除以 $e^x$ 可使一个隐式乘积变成 $y^2$，再求导。

As above, $y=1$ gives $e^x=2$.

Dividing the curve equation by $e^x$,

Differentiating,

At $y=1$ and $e^x=2$,

Hence $\boxed{m=-\frac54}$.