$$\begin{aligned} & 9 x^2+12 x+18 y-14=0 \\ & \left(x+\frac{2}{3}\right)^2=-2(y-1) \ldots(1) \end{aligned}$$

Equation of tangent to (1)

$$\begin{aligned} & t\left(x+\frac{2}{3}\right)=-(y-1)+\frac{1}{2} t^2 \text { passes through }(0,1) \\ & \Rightarrow \frac{2}{3} t=\frac{1}{2} t^2 \Rightarrow t=0, \frac{4}{3} \Rightarrow m=0, m=-\frac{4}{3} \end{aligned}$$

$$\begin{aligned} & \Rightarrow\left|\frac{m+\frac{4}{3}}{1-\frac{4}{3} m}\right|=|m| \Rightarrow\left|\frac{3 m+4}{3-4 m}\right|=|m| \\ & \Rightarrow 3 m+4=m(3-4 m) \text { or } 3 m+4=-m(3-4 m) \\ & 3 m+4=3 m-4 m^2 \text { or } 3 m+4=-3 m+4 m^2 \\ & 4 m^2+4=0 \text { (not possible) or } 4 m^2-6 m-4=0 \\ & m_1+m_2=\frac{3}{2}, m_1 m_2=-1 \\ & \Rightarrow m_1^2+m_2^2=\frac{17}{4} \\ & 16\left(m_1^2+m_2^2\right)=68 \end{aligned}$$