Step 1: Determine the required slope for the tangent lines.
The given line is $y = x$, which has slope 1. Since the tangent lines must be perpendicular to $y = x$, their slope is $-1$ (because $1 \times (-1) = -1$). So the tangents have the form $y = -x + c$, which can be written as $x + y - c = 0$.
Step 2: Substitute into the hyperbola equation.
The hyperbola is $x^2 - 2y^2 = 18$. Substituting $y = -x + c$: \[x^2 - 2(-x+c)^2 = 18\] \[x^2 - 2(x^2 - 2cx + c^2) = 18\] \[x^2 - 2x^2 + 4cx - 2c^2 = 18\] \[-x^2 + 4cx - 2c^2 - 18 = 0\] Multiplying through by $-1$: \[x^2 - 4cx + (2c^2 + 18) = 0\]
Step 3: Apply the tangency condition (discriminant = 0).
For the line to touch the hyperbola at exactly one point, the discriminant must be zero: \[D = (-4c)^2 - 4(1)(2c^2+18) = 0\] \[16c^2 - 8c^2 - 72 = 0 \implies 8c^2 = 72 \implies c^2 = 9 \implies c = \pm 3\]
Step 4: Write the two tangent lines.
The two tangent lines are $y = -x + 3$ and $y = -x - 3$, equivalently: $x + y - 3 = 0$ and $x + y + 3 = 0$. These are parallel lines (same coefficients of $x$ and $y$).
Step 5: Compute the distance between the two parallel lines.
Distance between $x + y + c_1 = 0$ and $x + y + c_2 = 0$ is $\dfrac{|c_1 - c_2|}{\sqrt{1^2+1^2}}$. With $c_1 = -3$ and $c_2 = 3$: \[d = \frac{|-3-3|}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}\]
Step 6: Verify numerically.
$3\sqrt{2} \approx 3 \times 1.414 = 4.24$. This is the distance between the two parallel tangent lines.
Step 7: State the final answer.
\[ \boxed{3\sqrt{2}} \]