一条曲线的方程为 $3x\ln y=x^2-4y^3$，其中 $x>0$。

求曲线上 $y=1$ 对应的点处的斜率。

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。

先用 $y=1$ 代回原方程求 $x$。再隐式求导，将所有含 $\mathrm{d}y/\mathrm{d}x$ 的项集中。

At $y=1$, the equation gives $x^2=4$. Since $x>0$, the point has $x=2$. Differentiating implicitly,

At $(2,1)$ this becomes $6y'=4-12y'$, giving the gradient $y'=2/9$.