If the general solution of the differential equation $$y' = \frac{y}{x} + \Phi\left(\frac{x}{y}\right)$$, for some function $$\Phi$$, is given by $$y\ln|cx| = x$$, where c is an arbitrary constant, then $$\Phi(2)$$ is equal to:

Let us denote $$z = \dfrac{x}{y}$$. Then $$y = \dfrac{x}{z}$$ and $$\dfrac{y}{x} = \dfrac{1}{z}\,.$$

The given general solution of the differential equation is $$y\ln|cx| = x \qquad -(1)$$ where $$c$$ is an arbitrary constant.

From $$(1)$$ we obtain $$\ln|cx| = \dfrac{x}{y} = z \qquad -(2)$$

Differentiate $$(1)$$ with respect to $$x$$:

$$\dfrac{d}{dx}\bigl(y\ln|cx|\bigr)=\dfrac{d}{dx}(x)$$ $$y' \ln|cx| + y\cdot\dfrac{1}{x} = 1$$

Hence $$y' = \dfrac{1 - \dfrac{y}{x}}{\ln|cx|} \qquad -(3)$$

Replace the fractions in $$(3)$$ using $$z = \dfrac{x}{y}$$ and $$(2)$$:

$$\dfrac{y}{x} = \dfrac{1}{z}, \qquad \ln|cx| = z$$

Therefore $$y' = \dfrac{1 - \dfrac{1}{z}}{z} = \dfrac{z - 1}{z^2} \qquad -(4)$$

The differential equation itself is $$y' = \dfrac{y}{x} + \Phi\!\left(\dfrac{x}{y}\right) = \dfrac{1}{z} + \Phi(z) \qquad -(5)$$

Equate $$(4)$$ and $$(5)$$:

$$\dfrac{1}{z} + \Phi(z) = \dfrac{z - 1}{z^2}$$

Solve for $$\Phi(z)$$:

$$\Phi(z) = \dfrac{z - 1}{z^2} - \dfrac{1}{z} = \dfrac{z - 1 - z}{z^2} = -\,\dfrac{1}{z^2} \qquad -(6)$$

Now evaluate at $$z = 2$$:

$$\Phi(2) = -\,\dfrac{1}{2^2} = -\,\dfrac{1}{4}$$

Hence the required value is $$-\dfrac{1}{4}$$, which matches Option D.