Verify that the given function $xy = \log y + C$ is a solution of the differential equation $y' = \frac{y^2}{1 - xy}$ $(xy \neq 1)$.

(A) Given function: $xy = \log y + C$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(xy) = \frac{d}{dx}(\log y + C)$
Using the product rule on the left side and the chain rule on the right side:
$y \cdot \frac{d}{dx}(x) + x \cdot \frac{dy}{dx} = \frac{1}{y} \cdot \frac{dy}{dx} + 0$
$y + xy' = \frac{1}{y} y'$
Multiply the entire equation by $y$ to eliminate the fraction:
$y^2 + xyy' = y'$
Rearrange the terms to isolate $y'$:
$y^2 = y' - xyy'$
$y^2 = y'(1 - xy)$
Thus,$y' = \frac{y^2}{1 - xy}$ (where $xy \neq 1$).
Since the derivative of the given function matches the differential equation,the function is indeed a solution.