Verify that the given function $x^{2}=2 y^{2} \log y$ is a solution of the corresponding differential equation $(x^{2}+y^{2}) \frac{dy}{dx}-xy=0$.

Given function: $x^{2}=2 y^{2} \log y$
Differentiating both sides with respect to $x$:
$2x = 2 \frac{d}{dx} [y^{2} \log y]$
$x = \frac{d}{dx} [y^{2} \log y]$
$x = 2y \log y \frac{dy}{dx} + y^{2} \cdot \frac{1}{y} \frac{dy}{dx}$
$x = \frac{dy}{dx} (2y \log y + y)$
$x = y \frac{dy}{dx} (2 \log y + 1)$
$\frac{dy}{dx} = \frac{x}{y(1+2 \log y)}$
Now,substitute $\frac{dy}{dx}$ into the $L.H.S.$ of the differential equation $(x^{2}+y^{2}) \frac{dy}{dx}-xy$:
$L.H.S. = (2y^{2} \log y + y^{2}) \cdot \frac{x}{y(1+2 \log y)} - xy$
$L.H.S. = y^{2}(2 \log y + 1) \cdot \frac{x}{y(1+2 \log y)} - xy$
$L.H.S. = y(2 \log y + 1) \cdot \frac{x}{(1+2 \log y)} - xy$
$L.H.S. = xy - xy = 0$
Since $L.H.S. = R.H.S.$,the given function is a solution of the differential equation.