Solution
= Solution
Use <implicit differentiation> on $F(x,y)=xy^3-yx^3-6=0$. Along the curve,
$$
y'=-\frac{F_x}{F_y}=-\frac{y^3-3yx^2}{3xy^2-x^3}.
$$
At $(1,2)$, $F_x=2$ and $F_y=11\ne0$, so the slope is $-2/11$. The <tangent line> is
$$
\boxed{y-2=-\frac2{11}(x-1),\quad\text{equivalently }2x+11y=24.}
$$