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 $y = a x + b$ where $a, b \in \mathbb{R}$ and $a \neq 0$. The $x$-coordinates of intersections of $d$ and $C$ are roots of
$$\sqrt[3]{x^2} = a x + b.$$
Let $t = \sqrt[3]{x}$, then $x = t^3$, so $\sqrt[3]{x^2} = t^2$. The equation becomes
$$t^2 = a t^3 + b \implies a t^3 - t^2 + b = 0.$$
This cubic equation $a t^3 - t^2 + b = 0$ has three distinct solutions $t_1, t_2, t_3$ and $t_1 t_2 t_3 \neq 0$ (because $b \neq 0$). 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
$$(t_1 t_2)^3 + (t_2 t_3)^3 + (t_3 t_1)^3 = 3 (t_1 t_2)(t_2 t_3)(t_3 t_1) = 3 t_1^2 t_2^2 t_3^2.$$
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,
$$\sqrt[3]{\frac{x_1^2}{x_2 x_3}} + \sqrt[3]{\frac{x_2^2}{x_3 x_1}} + \sqrt[3]{\frac{x_3^2}{x_1 x_2}} = \frac{t_1^2}{t_2 t_3} + \frac{t_2^2}{t_3 t_1} + \frac{t_3^2}{t_1 t_2} \\ = - (t_1 + t_2) \left( \frac{t_1}{t_2^2} + \frac{t_2}{t_1^2} \right) + \frac{t_1 t_2}{(t_1 + t_2)^2} \\ = - \left( \frac{t_1^2}{t_2^2} + \frac{t_2^2}{t_1^2} + \frac{t_1}{t_2} + \frac{t_2}{t_1} \right) + \frac{t_1 t_2}{(t_1 + t_2)^2}.$$
Using the AM-GM inequality, we have
$$\frac{t_1^2}{t_2^2} + \frac{t_2^2}{t_1^2} \geq 2, \quad \frac{t_1}{t_2} + \frac{t_2}{t_1} \geq 2, \quad \frac{t_1 t_2}{(t_1 + t_2)^2} \leq \frac{1}{4}.$$
Therefore,
$$\sqrt[3]{\frac{x_1^2}{x_2 x_3}} + \sqrt[3]{\frac{x_2^2}{x_3 x_1}} + \sqrt[3]{\frac{x_3^2}{x_1 x_2}} \leq - (2 + 2) + \frac{1}{4} = -\frac{15}{4}.$$
Note that $t_1, t_2$ are distinct so the equality does not occur.