If the general solution of \((\mathrm{dy} / \mathrm{dx})=(\mathrm{y} / \mathrm{x})+\mathrm{f}(\mathrm{x} / \mathrm{y})\) is \(\mathrm{y}\) \(=[\mathrm{x} / \log |\mathrm{cx}|]\), then \(\mathrm{f}(\mathrm{x} / \mathrm{y})\) is given by: (A) \(\left(\mathrm{x}^{2} / \mathrm{y}^{2}\right)\) (B) \(\left(\mathrm{y}^{2} / \mathrm{x}^{2}\right)\) (C) \(\left[\left(-\mathrm{x}^{2}\right) / \mathrm{y}^{2}\right]\) (D) \(\left[\left(-\mathrm{y}^{2}\right) / \mathrm{x}^{2}\right]\)

We are given the function \(y = \dfrac{x}{\log|cx|}\). Differentiate with respect to x to find the derivative, \(\frac{dy}{dx}\). \[ \begin{aligned} \frac{dy}{dx} &= \frac{d}{dx} \left( \frac{x}{\log|cx|} \right) \\ &= \frac{1}{\log|cx|} + x \cdot \frac{d}{dx} \left( \frac{1}{\log|cx|} \right) \end{aligned} \] Using chain rule, \[ \begin{aligned} \frac{d}{dx}(\log|x|) &= \frac{1}{x} \\ \frac{d}{dx}(\log|cx|) &= \frac{d}{dx}( \log(c) + \log|x|) = \frac{1}{x} \end{aligned} \] Now, we can continue differentiating \(\frac{dy}{dx}\): \[ \begin{aligned} \frac{dy}{dx} &= \frac{1}{\log|cx|} + x \cdot \frac{-1}{x \log^2|cx|} \\ &= \frac{1}{\log|cx|} - \frac{1}{\log^2|cx|} \end{aligned} \]

Substitute the expression of \(\frac{dy}{dx}\) and \(y\) we found in step 1 into the given differential equation. Let \(z = \frac{x}{y}\), we need to find \(f(z) = f \left(\frac{x}{y} \right)\): \[ \begin{aligned} \frac{1}{\log|cx|} - \frac{1}{\log^2|cx|} &= \frac{x}{log |cx|} + f \left(\frac{x}{y} \right) \end{aligned} \] Subtract \(\frac{x}{\log |cx|}\) from both sides and substitute \(z = \frac{x}{y}\) to get: \[ \begin{aligned} f(z) = \frac{1}{\log|cx|} - \frac{1}{\log^2|cx|} - \frac{x}{\log |cx|} \end{aligned} \] Now, eliminate x from the equation by dividing both sides by z which is \(\frac{x}{y}\): \[ \begin{aligned} f(z) &= \frac{1}{\log|cx|} \cdot \frac{y}{x} - \frac{1}{\log^2|cx|} \cdot \frac{y}{x} - \frac{y}{\log |cx|} \\ &= \frac{y^2}{x^2} - \frac{y^2}{x^2\log|cx|} \end{aligned} \] Combine the terms: \[ \begin{aligned} f(z) &= \frac{-y^2}{x^2\log|cx|} \end{aligned} \] Since \(z = \frac{x}{y}\), we can rewrite the function \(f(z)\) as: \[ \begin{aligned} f(z) = -\frac{1}{z^2\log|cz|} \end{aligned} \] So we find that the function f(x/y) is given by option (D): \[ \begin{aligned} f \left(\frac{x}{y}\right) = \left[\frac{-y^2}{x^2}\right] \end{aligned} \]