Let a smooth curve $$y = f(x)$$ be such that the slope of the tangent at any point $$(x, y)$$ on it is directly proportional to $$\left(\dfrac{-y}{x}\right)$$. If the curve passes through the points $$(1, 2)$$ and $$(8, 1)$$, then $$\left|y\left(\dfrac{1}{8}\right)\right|$$ is equal to

The slope of the tangent at any point $$(x, y)$$ is directly proportional to $$\left(\dfrac{-y}{x}\right)$$, so we have:

$$ \frac{dy}{dx} = k\left(\frac{-y}{x}\right) $$

Separating variables yields:

$$ \frac{dy}{y} = -k\frac{dx}{x} $$

Integrating both sides gives:

$$ \ln|y| = -k\ln|x| + C $$

and exponentiating leads to:

$$ y = Ax^{-k} $$

Substituting the point $$(1, 2)$$ into the expression $$y = Ax^{-k}$$ gives:

$$ 2 = A \cdot 1^{-k} \implies A = 2 $$

Hence the solution simplifies to:

$$ y = 2x^{-k} $$

Applying the condition $$(8, 1)$$ next leads to:

$$ 1 = 2 \cdot 8^{-k} \implies 8^{-k} = \frac{1}{2} \implies 8^k = 2 \implies k = \frac{1}{3} $$

Therefore:

$$ y = 2x^{-1/3} $$

Finally, to find $$\left|y\left(\frac{1}{8}\right)\right|$$, we compute:

$$ y\left(\frac{1}{8}\right) = 2\left(\frac{1}{8}\right)^{-1/3} = 2 \times 8^{1/3} = 2 \times 2 = 4 $$

The answer is Option B: $$4$$.