Minimum Problem

Since $y$ is an even function, it is symmetrical on the $y$ -axis. It has two asymptotes at $x=\pm 1$ . The shortest distances are the normals from the three parts of the curve $y$ that passes through $P(0,1)$ . Let the gradient of a line passing through $P(0,1)$ be $m= \dfrac {y-1}x$ . Then, the normal will also have a gradient of $m$ . Let us find the gradient of a point $(x,y)$ of the curve.

$\begin{aligned} y & = \frac {x^2+1}{x^2-1} \\ (x^2-1)y & = x^2+1 & \small \color{#3D99F6} \text{Differentiating both sides w.r.t }x \\ 2xy +(x^2-1)\frac {dy}{dx} & = 2x \\ (x^2-1)\frac {dy}{dx} & = 2x(1-y) \\ & = 2x \left(1- \frac {x^2+1}{x^2-1}\right) \\ & = \frac {-4x}{x^2-1} \\ \frac {dy}{dx} & = \frac {-4x}{(x^2-1)^2} & \small \color{#3D99F6} \text{Note that } \frac {dy}{dx} = - \frac 1m \\ \implies \frac {-4x}{(x^2-1)^2} & = - \frac x{y-1} \\ 4x(y-1) - x(x^2-1)^2 & = 0 \\ \frac {8x}{x^2-1} - x(x^2-1)^2 & = 0 \\ 8x - x(x^2-1)^3 & = 0 \\ x \left(8-(x^2-1)^3 \right) & = 0 \end{aligned}$

$\implies x = \begin{cases} x = 0 \\ (x^2-1)^3 = 8 & \implies x = \pm \sqrt 3 \end{cases}$

$\implies \left(3x_1 + 6x_2 + 9x_3\right)^2 = \left(-3\sqrt 3 + 6(0) + 9\sqrt 3\right)^2 = \boxed{108}$