VMO

a. It is easy to see that $d$ cannot be a line parallel to the $y$-axis (otherwise $d$ only intersects $C$ at at most one point), so $d$ has the form $y=ax+b$ where $a,b\in \mathbb{R}$ and $a\ne 0$. The $x$-coordinates of intersections of $d$ and $C$ are roots of $$\sqrt[3]{x^2}=ax+b.$$ Let $t=\sqrt[3]{x}$, then $x=t^3$, so $\sqrt[3]{x^2}=t^2$. The equation becomes $$t^2=a{t}^3+b⟹a{t}^3-t^2+b=0.$$ This cubic equation $a{t}^3-t^2+b=0$ has three distinct solutions $t_1,t_2,t_3$ and $t_1{t}_2{t}_3\ne 0$ (because $b\ne 0$). Applying Vieta's theorem to the cubic equation, we have $t_1{t}_2+t_2{t}_3+t_3{t}_1=0$, which implies $$(t_1{t}_2{)}^3+(t_2{t}_3{)}^3+(t_3{t}_1{)}^3=3(t_1{t}_2)(t_2{t}_3)(t_3{t}_1)=3t_1^2{t}_2^2{t}_3^2.$$ We can rewrite this equality as $$\frac{t_2{t}_3}{t_1^2}+\frac{t_3{t}_1}{t_2^2}+\frac{t_1{t}_2}{t_3^2}=3,$$ or $$\sqrt[3]{\frac{x_2{x}_3}{x_1^2}}+\sqrt[3]{\frac{x_3{x}_1}{x_2^2}}+\sqrt[3]{\frac{x_1{x}_2}{x_3^2}}=3.$$ So $\sqrt[3]{\frac{x_2{x}_3}{x_1^2}}+\sqrt[3]{\frac{x_3{x}_1}{x_2^2}}+\sqrt[3]{\frac{x_1{x}_2}{x_3^2}}=3$ is a constant.

b. Without loss of generality, assume that $t_1$ and $t_2$ have the same signs. From $t_1{t}_2+t_2{t}_3+t_3{t}_1=0$, we have $t_3=-\frac{t_1{t}_2}{t_1+t_2}$. Hence, $$\begin{gathered} \sqrt[3]{\frac{x_1^2}{x_2{x}_3}}+\sqrt[3]{\frac{x_2^2}{x_3{x}_1}}+\sqrt[3]{\frac{x_3^2}{x_1{x}_2}}=\frac{t_1^2}{t_2{t}_3}+\frac{t_2^2}{t_3{t}_1}+\frac{t_3^2}{t_1{t}_2} \\ =-(t_1+t_2)\left(\frac{t_1}{t_2^2}+\frac{t_2}{t_1^2}\right)+\frac{t_1{t}_2}{(t_1+t_2{)}^2} \\ =-\left(\frac{t_1^2}{t_2^2}+\frac{t_2^2}{t_1^2}+\frac{t_1}{t_2}+\frac{t_2}{t_1}\right)+\frac{t_1{t}_2}{(t_1+t_2{)}^2}. \end{gathered}$$ Using the AM-GM inequality, we have $$\frac{t_1^2}{t_2^2}+\frac{t_2^2}{t_1^2}\ge 2,\frac{t_1}{t_2}+\frac{t_2}{t_1}\ge 2,\frac{t_1{t}_2}{(t_1+t_2{)}^2}\le \frac{1}{4}.$$ Therefore, $$\sqrt[3]{\frac{x_1^2}{x_2{x}_3}}+\sqrt[3]{\frac{x_2^2}{x_3{x}_1}}+\sqrt[3]{\frac{x_3^2}{x_1{x}_2}}\le -(2+2)+\frac{1}{4}=-\frac{15}{4}.$$ Note that $t_1,t_2$ are distinct so the equality does not occur.