1. Write the parabola in standard form

Given: $9x^2+12x+18y-14=0$

Rearrange: $18y=-9x^2-12x+14$ $y=-\frac{1}{2}x^2-\frac{2}{3}x+\frac{7}{9}$

Complete the square in $x$: $9x^2+12x+18y-14=0$ $9\left(x^2+\frac{4}{3}x\right)+18y-14=0$ $9\left[\left(x+\frac{2}{3}\right)^2-\frac{4}{9}\right]+18y-14=0$ $9\left(x+\frac{2}{3}\right)^2-4+18y-14=0$ $9\left(x+\frac{2}{3}\right)^2+18y-18=0$ $\left(x+\frac{2}{3}\right)^2=-2(y-1)$

So the parabola is $\boxed{(x+\tfrac23)^2=-2(y-1)}$ which has vertex $V\left(-\frac23,1\right)$ and parameter $a=-\frac12$.

2. Find tangents from $P(0,1)$ to the parabola

A line through $P(0,1)$ with slope $m$ is $y=mx+1.$

Substitute into the parabola: $9x^2+12x+18(mx+1)-14=0$ $9x^2+(12+18m)x+4=0.$

For tangency, discriminant must be zero: $(12+18m)^2-4\cdot 9\cdot 4=0$ $(12+18m)^2=144$ $12+18m=\pm 12.$

Hence,

• if $12+18m=12$, then $m=0$;
• if $12+18m=-12$, then $18m=-24$, so $m=-\frac43$.

Thus the two tangents are $L_1:y=1, \qquad L_2:y=-\frac43x+1.$

3. Use the isosceles condition

$Q$ lies on $L_1$ and $R$ lies on $L_2$, and $\triangle PQR$ is isosceles with base $QR$.

Since base is $QR$, the equal sides are $PQ=PR.$

So $Q$ and $R$ must be at equal distance from $P$ on the two tangent lines.

Let $Q=(a,1)$ on $L_1$. Then $PQ=|a|.$

Let $R$ be on $L_2$. A convenient parametrization is $R=(t,1-\tfrac43 t).$ Then $PR=\sqrt{t^2+\left(-\frac43 t\right)^2}=\frac53|t|.$

Condition $PQ=PR$ gives $|a|=\frac53|t|.$

Because there are two possible directions on each line, this gives the two cases $a=\frac53 t \quad \text{or} \quad a=-\frac53 t.$

4. Find slope of $QR$ in each case

The slope of $QR$ is $m=\frac{\left(1-\frac43 t\right)-1}{t-a}=\frac{-\frac43 t}{t-a}.$

Case 1: $a=\frac53 t$

Then $t-a=t-\frac53 t=-\frac23 t,$ so $m=\frac{-\frac43 t}{-\frac23 t}=2.$

Case 2: $a=-\frac53 t$

Then $t-a=t+\frac53 t=\frac83 t,$ so $m=\frac{-\frac43 t}{\frac83 t}=-\frac12.$

Hence, $m_1=2, \qquad m_2=-\frac12.$

5. Compute the required value

$m_1^2+m_2^2=2^2+\left(-\frac12\right)^2=4+\frac14=\frac{17}{4}.$

Therefore, $16(m_1^2+m_2^2)=16\cdot \frac{17}{4}=68.$

So the required integer is $\boxed{68}.$